Free math worksheets

- Recall all three ratio-test outcomes, including the equality case. - Distinguish “the test is inconclusive” from “the series diverges.” - Ask what additional method would be needed after a boundary result.

1. The ratio test proves absolute convergence when \(L<1\) and divergence when \(L>1\). 2. When \(L=1\), the test gives no conclusion. 3. Therefore the student’s conclusion is invalid, and another test is required.

The conclusion is invalid; \(L=1\) makes the ratio test inconclusive.

- Compare \(0.999\) with the exact boundary value \(1\). - The distance from \(1\) does not matter if the inequality is strict. - State the stronger convergence conclusion supplied by the ratio test.

1. The value \(0.999\) is strictly less than \(1\). 2. Any ratio-test limit below \(1\) proves absolute convergence, even when it is close to \(1\).

The series converges absolutely.

- Compare an unbounded ratio limit with \(1\). - Identify which decisive ratio-test case applies. - Do not treat \(+\infty\) as an inconclusive limit.

1. The limit \(+\infty\) is greater than \(1\). 2. Therefore the series diverges by the ratio test.

The series diverges by the ratio test.

- Write \(a_{n+1}\) by replacing \(n\) with \(n+1\). - Use \((n+1)!=(n+1)n!\) before taking the limit. - Compare the resulting limit with the ratio-test boundaries \(0\) and \(1\).

1. For \(a_n=\frac{3^n}{n!}\), \(\left|\frac{a_{n+1}}{a_n}\right|=\frac{3^{n+1}}{(n+1)!}\frac{n!}{3^n}=\frac3{n+1}\). 2. Therefore \(L=\lim_{n\to\infty}\frac3{n+1}=0<1\). 3. The series converges absolutely by the ratio test.

\(L=0\), so the series converges absolutely by the ratio test.

- Write \((n+1)!=(n+1)n!\) in the consecutive-term quotient. - Cancel the common factorial and power factors before taking the limit. - Compare the resulting limit with \(1\).

1. For \(a_n=\frac{n!}{3^n}\), \(\left|\frac{a_{n+1}}{a_n}\right|=\frac{(n+1)!}{3^{n+1}}\frac{3^n}{n!}=\frac{n+1}{3}\). 2. Thus \(L=\lim_{n\to\infty}\frac{n+1}{3}=\infty>1\). 3. Therefore the series diverges by the ratio test.

\(L=\infty\), so the series diverges by the ratio test.

- Write \(a_{n+1}\) by increasing every occurrence of \(n\) by \(1\). - Multiply by the reciprocal of \(a_n\) before simplifying. - A ratio limit of \(1\) gives no convergence conclusion.

1. For \(a_n=\frac{n+1}{2n+3}\), \(\left|\frac{a_{n+1}}{a_n}\right|=\frac{n+2}{2n+5}\frac{2n+3}{n+1}\). 2. Each rational factor has a finite nonzero limit, and their product approaches \(1\). 3. Thus \(L=1\), so the ratio test is inconclusive.

\(L=1\), so the ratio test is inconclusive.

- Form the quotient of the two reciprocal squares carefully. - Rewrite the result as \(\left(\frac{n}{n+1}\right)^2\). - State the ratio test’s exact conclusion at \(L=1\), regardless of what another test says.

1. For \(a_n=\frac1{n^2}\), \(\left|\frac{a_{n+1}}{a_n}\right|=\frac{n^2}{(n+1)^2\!}\). 2. Therefore \(L=\lim_{n\to\infty}\left(\frac{n}{n+1}\right)^2=1\). 3. The ratio test is inconclusive; it does not prove convergence or divergence for this series.

\(L=1\), so the ratio test proves neither convergence nor divergence.

- Absolute values remove the alternating sign from the powers of \(-3\). - Use \((n+1)!=(n+1)n!\) to simplify the quotient. - Compare the resulting limit with \(1\).

1. For \(a_n=\frac{(-3)^n}{n!}\), \(\left|\frac{a_{n+1}}{a_n}\right|=\frac3{n+1}\). 2. Thus \(L=0<1\). 3. Therefore the series converges absolutely by the ratio test.

\(L=0\), so the series converges absolutely by the ratio test.

- Treat \(c\) as fixed when simplifying the consecutive-term quotient. - Use \((n+1)!=(n+1)n!\). - Compare the unbounded ratio with the ratio-test threshold \(1\).

1. For \(a_n=\frac{n!}{c^n}\), \(\left|\frac{a_{n+1}}{a_n}\right|=\frac{n+1}{c}\). 2. For every fixed \(c>0\), the ratio tends to \(\infty\), so \(L>1\). 3. Therefore the series diverges by the ratio test for every \(c>0\).

\(L=\infty\) for every \(c>0\), so the series diverges by the ratio test.

- The consecutive-term ratio is already supplied, so no reconstruction of \(a_n\) is needed. - Use the leading coefficients to evaluate its limit. - Compare that limit directly with \(1\).

1. The ratio limit is \(L=\lim_{n\to\infty}\frac{n+2}{2n+1}=\frac12\). 2. Since \(L<1\), the positive-term series converges by the ratio test.

\(L=\frac12\), so the series converges by the ratio test.

- Take the limit of the supplied ratio without trying to recover the terms. - Track what happens to \(2/n\) as \(n\to\infty\). - Recall that equality at the ratio-test boundary gives no conclusion.

1. The ratio limit is \(L=\lim_{n\to\infty}\left(1+\frac2n\right)=1\). 2. The ratio test is inconclusive when \(L=1\).

\(L=1\), so the ratio test is inconclusive.

- Determine the sign of the raw quotient before taking absolute values. - Recall that the ratio test is formulated for magnitudes. - Simplify the factorial quotient after removing the sign.

1. The ratio test compares consecutive term magnitudes. 2. Without absolute values, the consecutive ratios are negative and are not the quantities used by the test. 3. With absolute values, \(\left|a_{n+1}/a_n\right|=\frac3{n+1}\to0\), proving absolute convergence.

Absolute values are required because the ratio test analyzes term magnitudes; here they give a ratio limit of \(0\).

- Examine the even- and odd-indexed ratio subsequences separately. - Check whether they approach the same value. - The standard limit form of the ratio test needs one ratio limit.

1. The standard ratio test in limit form requires \(\lim|a_{n+1}/a_n|\) to exist. 2. Here the ratio sequence has two different subsequential values, so that limit does not exist. 3. Therefore the standard ratio test is inconclusive.

The standard ratio test is inconclusive because the required ratio limit does not exist.

- Separate the first \(19\) terms from the remaining infinite tail. - Recall whether adding a finite sum changes convergence. - Apply the ratio test only where the quotient calculation is valid.

1. Convergence depends on the behavior of an infinite tail. 2. The ratio test may be applied to the tail beginning at \(n=20\). 3. The finitely many preceding terms do not affect whether the full series converges or diverges.

No. The ratio test can be applied from \(n=20\) onward because finitely many initial terms do not affect convergence.

- Compare \((n+1)!\) with \(n!\). - Compare \(c^{n+1}\) with \(c^n\) for a fixed base \(c\). - Consider why these cancellations are more direct than analyzing the original terms.

1. Consecutive factorials cancel to a small number of new linear factors. 2. Consecutive exponential powers reduce to a constant base factor. 3. The resulting quotient is usually straightforward to simplify and compare with \(1\).

Consecutive-term ratios simplify factorials and exponential powers directly, often leaving an elementary limit.

- Decide whether signs or magnitudes belong in the consecutive-term comparison. - Recompute the comparison in the required nonnegative form. - Distinguish an incorrect justification from an incorrect conclusion.

1. The ratio test uses \(|a_{n+1}/a_n|\), not the signed ratio. 2. Here \(|a_{n+1}/a_n|=1/2\). 3. Since \(1/2<1\), the series of absolute values converges. 4. Therefore the original series converges absolutely. The conclusion was correct, but the student's criterion was stated incorrectly.

The correct ratio is \(1/2\), so the series converges absolutely.

- Compare the newly introduced factorial factor with the new exponential factor. - Determine how their quotient behaves as the index grows. - Connect a limiting quotient above the boundary with divergence.

1. Let \(a_n=n!/4^n\). 2. The consecutive-term ratio is \(a_{n+1}/a_n=(n+1)/4\). 3. This ratio tends to infinity, which is greater than \(1\). 4. Therefore the series diverges by the ratio test.

The series diverges.

- Compare the polynomial factor at consecutive indices. - Separate its ratio from the constant exponential ratio. - Evaluate the product of the limiting factors.

1. Let \(a_n=n^2/2^n\). 2. The consecutive-term ratio is \(a_{n+1}/a_n=\frac12((n+1)/n)^2\). 3. The ratio approaches \(1/2<1\). 4. Therefore the series converges by the ratio test.

The series converges.

- Compute the consecutive quotient for each example separately. - Compare the two limiting quotients with the known behavior of the original series. - Can the shared boundary value force a unique conclusion?

1. For \(a_n=1/n\), \(\left|\frac{a_{n+1}}{a_n}\right|=\frac{n}{n+1}\to1\). 2. For \(b_n=1/n^2\), \(\left|\frac{b_{n+1}}{b_n}\right|=\left(\frac{n}{n+1}\right)^2\to1\). 3. The harmonic series diverges, while the \(p\)-series with \(p=2\) converges. 4. Therefore a ratio-test limit of \(1\) gives no conclusion about convergence or divergence.

Both ratio-test limits equal \(1\). The first series diverges and the second converges, so the ratio test is inconclusive when its limit is \(1\).

- Write the consecutive-term ratio for the reciprocal sequence. - Use the provided limit in the reciprocal direction. - Compare the resulting constant with the ratio-test boundary.

1. Let \(a_n=1/F_n\). 2. The consecutive-term ratio is \(\frac{a_{n+1}}{a_n}=\frac{F_n}{F_{n+1}}\). 3. Using the given Fibonacci ratio limit, \(\frac{F_n}{F_{n+1}}\to\frac{1}{\varphi}\). 4. Since \(\varphi>1\), \(1/\varphi<1\). 5. Therefore \(\sum1/F_n\) converges by the ratio test.

The series converges by the ratio test; the ratio-test limit is \(\frac{1}{\varphi}=\frac{\sqrt5-1}{2}\).

- The sign in the recurrence disappears when testing absolute convergence. - Express the test directly through the given consecutive-term relationship. - Compare the leading coefficients of the resulting rational expression.

1. Taking absolute values in the recurrence gives \(\left|\frac{a_{n+1}}{a_n}\right|=\frac{n+2}{4n+1}\). 2. Dividing numerator and denominator by \(n\) gives the limit \(1/4\). 3. Since \(1/4<1\), the series of absolute values converges by the ratio test. 4. Therefore \(\sum a_n\) converges absolutely.

The series converges absolutely by the ratio test; the absolute-ratio limit is \(\frac14\).

- Form the quotient using both the cubic factor and the power of \(2\). - Rewrite \(\frac{(n+1)^3}{n^3}\) as \(\left(1+\frac1n\right)^3\). - Compare the final limit with \(1\) and state whether the conclusion is absolute convergence.

1. For \(a_n=\frac{n^3}{2^n}\), \(\left|\frac{a_{n+1}}{a_n}\right|=\left(\frac{n+1}{n}\right)^3\frac12\). 2. Since \(\frac{n+1}{n}\to1\), the ratio-test limit is \(L=\frac12<1\). 3. Therefore the series converges absolutely by the ratio test.

\(L=\frac12\), so the series converges absolutely by the ratio test.

- Replacing \(n\) by \(n+1\) multiplies the exponential factor by \(2\). - Keep the polynomial change as the quotient \(n^3/(n+1)^3\). - Take the limit only after combining both factors.

1. For \(a_n=\frac{2^n}{n^3}\), \(\left|\frac{a_{n+1}}{a_n}\right|=\frac{2^{n+1}}{(n+1)^3}\frac{n^3}{2^n}=2\left(\frac{n}{n+1}\right)^3\). 2. Since \(\frac{n}{n+1}\to1\), the ratio-test limit is \(L=2>1\). 3. Therefore the series diverges by the ratio test.

\(L=2\), so the series diverges by the ratio test.

- Expand \((n+1)!\) by one factor and \((2n+2)!\) by two factors. - Cancel \(n!\) and \((2n)!\) before simplifying the quotient. - Determine the limit from the remaining polynomial factors.

1. For \(a_n=\frac{n!}{(2n)!}\), \(\left|\frac{a_{n+1}}{a_n}\right|=\frac{(n+1)!}{(2n+2)!}\frac{(2n)!}{n!}=\frac{n+1}{(2n+2)(2n+1)}\). 2. The quotient approaches \(0\), so \(L=0<1\). 3. Therefore the series converges absolutely by the ratio test.

\(L=0\), so the series converges absolutely by the ratio test.

- Expand each copy of \((n+1)!\) and the two new factors in \((2n+2)!\). - Cancel the common factorials before taking the limit. - Compare the leading coefficients of the remaining quadratic factors.

1. For \(a_n=\frac{(n!)^2}{(2n)!}\), \(\left|\frac{a_{n+1}}{a_n}\right|=\frac{((n+1)!)^2}{(2n+2)!}\frac{(2n)!}{(n!)^2}=\frac{(n+1)^2}{(2n+2)(2n+1)}\). 2. Dividing numerator and denominator by \(n^2\) gives \(L=\frac14<1\). 3. Therefore the series converges absolutely by the ratio test.

\(L=\frac14\), so the series converges absolutely by the ratio test.

- Form the quotient of the fifth-power factors and the powers of \(5\) separately. - Determine the limit of \(\left(1+\frac1n\right)^5\). - A ratio limit below \(1\) confirms exponential domination.

1. For \(a_n=\frac{n^5}{5^n}\), \(\left|\frac{a_{n+1}}{a_n}\right|=\left(\frac{n+1}{n}\right)^5\frac15\). 2. Since \(\frac{n+1}{n}\to1\), the ratio-test limit is \(L=\frac15<1\). 3. Therefore the series converges absolutely by the ratio test, so the exponential denominator dominates the polynomial numerator.

Yes. \(L=\frac15<1\), so the series converges absolutely by the ratio test; the exponential denominator dominates the polynomial numerator.

- The exponential factor gains one factor of \(5\) from one term to the next. - Express the polynomial contribution as \(\left(\frac{n}{n+1}\right)^5\). - Combine the two limiting factors before applying the ratio test.

1. For \(a_n=\frac{5^n}{n^5}\), \(\left|\frac{a_{n+1}}{a_n}\right|=5\left(\frac{n}{n+1}\right)^5\). 2. Since \(\frac{n}{n+1}\to1\), the ratio-test limit is \(L=5>1\). 3. Therefore the series diverges by the ratio test.

The limiting factor is \(L=5\), so the series diverges by the ratio test.

- Handle \(c=0\) separately before canceling powers of \(c\). - For \(c\ne0\), treat \(c\) as fixed and take absolute values in the consecutive-term quotient. - Determine whether the resulting limit depends on the size of the nonzero parameter.

1. If \(c=0\), the series is \(1+0+0+\cdots\), so it converges absolutely. 2. If \(c\ne0\), then for \(a_n=\frac{c^n}{n!}\), \(\left|\frac{a_{n+1}}{a_n}\right|=\frac{|c|}{n+1}\to0<1\). 3. Therefore the series converges absolutely for every real \(c\).

For \(c\ne0\), the ratio-test limit is \(L=0\); for \(c=0\), the series is \(1+0+0+\cdots\). Thus the series converges absolutely for every real \(c\).

- Treat \(c=0\) separately before forming a quotient that cancels powers of \(c\). - For \(c\ne0\), separate the polynomial quotient from the constant power ratio. - Compare \(|c|/5\) with \(1\) and preserve the boundary case.

1. If \(c=0\), every term is \(0\), so the series converges absolutely. 2. If \(c\ne0\), then for \(a_n=n^2\left(\frac{c}{5}\right)^n\), \(\left|\frac{a_{n+1}}{a_n}\right|=\left(\frac{n+1}{n}\right)^2\frac{|c|}{5}\to\frac{|c|}{5}\). 3. The ratio test proves absolute convergence when \(0<|c|<5\) and divergence when \(|c|>5\). 4. When \(|c|=5\), the ratio test is inconclusive.

The series converges absolutely for \(|c|<5\) (directly when \(c=0\), and by the ratio test when \(0<|c|<5\)), diverges for \(|c|>5\), and the ratio test is inconclusive for \(|c|=5\).

- Replace the varying ratio bound by a fixed number below one on a tail. - Iterate the fixed bound to compare term magnitudes with a geometric sequence. - Add the finite initial terms after controlling the tail.

1. For \(n\ge2\), \(1/(n+2)\le1/4\), so \(|a_{n+1}/a_n|\le3/4\). 2. Repeatedly applying the bound gives \(|a_{2+k}|\le|a_2|(3/4)^k\). 3. The magnitude tail is bounded by a convergent geometric series. 4. Therefore \(\sum|a_n|\) converges, so \(\sum a_n\) converges absolutely.

The series converges absolutely.

- Express each new consecutive-term ratio using the two ratios already given. - Treat multiplication and division separately. - Compare each resulting limit with the ratio-test boundary.

1. For the product terms, \(\left|\frac{a_{n+1}b_{n+1}}{a_nb_n}\right| =\left|\frac{a_{n+1}}{a_n}\right|\left|\frac{b_{n+1}}{b_n}\right| \to\frac23\cdot\frac34=\frac12\). 2. Since \(\frac12<1\), \(\sum a_nb_n\) converges absolutely. 3. For the quotient terms, \(\left|\frac{a_{n+1}/b_{n+1}}{a_n/b_n}\right| =\left|\frac{a_{n+1}}{a_n}\right|\left|\frac{b_n}{b_{n+1}}\right| \to\frac{2/3}{3/4}=\frac89\). 4. Since \(\frac89<1\), \(\sum a_n/b_n\) also converges absolutely.

Both series converge absolutely. Their ratio-test limits are \(\frac12\) for \(\sum a_nb_n\) and \(\frac89\) for \(\sum a_n/b_n\).

- Focus on the consecutive changes in the term magnitudes. - Separate the cases in which the limiting change lies below, at, or above the boundary value. - Compare what can be proved about magnitudes with what conditional convergence would require.

1. If \(L<1\), the ratio test proves convergence of \(\sum|a_n|\), so the series is absolutely convergent. 2. If \(L>1\) or the ratios grow without bound, the terms do not approach \(0\), so the series diverges. 3. If \(L=1\), the ratio test is inconclusive. 4. Therefore the ratio test never directly proves conditional convergence.

No. The ratio test proves absolute convergence when \(L<1\), divergence when \(L>1\), and is inconclusive when \(L=1\).

- Turn the consecutive-term bound into a bound for every later term. - Compare the absolute tail with a geometric tail. - Use the first term of that tail rather than the last included term.

1. The ratio bound gives \(|a_{21}|\le0.4|a_{20}|=2\times10^{-6}\). 2. Repeatedly applying the bound gives \(|a_{21+j}|\le(2\times10^{-6})(0.4)^j\) for \(j\ge0\). 3. Therefore \(\left|\sum_{n=21}^{\infty}a_n\right| \le\sum_{j=0}^{\infty}(2\times10^{-6})(0.4)^j =\frac{2\times10^{-6}}{1-0.4}\). 4. The bound equals \(\frac{1}{300000}\).

\(\left|\sum_{n=21}^{\infty}a_n\right|\le\frac{1}{300000}\).

- Subtract the two consecutive square exponents before simplifying the ratio. - Separate the cases according to the size of the parameter’s magnitude. - Compare exponential growth or decay with the remaining linear factor.

1. If \(c=0\), every term is \(0\), so the series converges absolutely. 2. For \(c\ne0\), let \(a_n=c^{n^2}/n!\). Then \(\left|\frac{a_{n+1}}{a_n}\right|=\frac{|c|^{(n+1)^2-n^2}}{n+1}=\frac{|c|^{2n+1}}{n+1}\). 3. If \(0<|c|<1\), the ratio tends to \(0\). If \(|c|=1\), it also tends to \(0\). 4. If \(|c|>1\), the exponential numerator grows without bound relative to \(n+1\), so the ratio tends to infinity. 5. Therefore the series converges absolutely for \(|c|\le1\) and diverges for \(|c|>1\).

The series converges absolutely for \(|c|\le1\) and diverges for \(|c|>1\).

- Compare consecutive combinations before expanding factorials fully. - Account for the additional factor in the exponential denominator. - Evaluate the rational ratio limit.

1. Let \(a_n=\binom n4/5^n\). 2. Using \(\binom{n+1}{4}/\binom n4=(n+1)/(n-3)\), \(|a_{n+1}/a_n|=\frac{n+1}{5(n-3)}\). 3. The ratio approaches \(1/5<1\). 4. Therefore the series converges by the ratio test.

The series converges.

- Separate the polynomial factor from the repeated parameter power in the ratio. - Use the test where the limiting ratio is strictly below or above one. - Examine the term limit at the boundary.

1. Let \(a_n=n^3x^n\). Then \(|a_{n+1}/a_n|=|x|((n+1)/n)^3\to|x|\). 2. The series converges absolutely when \(|x|<1\) and diverges when \(|x|>1\). 3. If \(x=1\), the terms are \(n^3\), and if \(x=-1\), their magnitudes are \(n^3\). 4. At both boundary values the terms fail to approach \(0\), so the series diverges.

The series converges absolutely for \(|x|<1\) and diverges for \(|x|\ge1\).

- Read the consecutive change directly from the recurrence. - Expand several recurrence steps and look for cancellation. - Investigate the series directly when the limiting change reaches the boundary value.

1. The consecutive-term ratio is \(a_{n+1}/a_n=n/(n+1)\to1\), so the ratio test is inconclusive. 2. Repeated substitution gives \(a_n=(1/2)(2/3)\cdots((n-1)/n)=1/n\). 3. Therefore \(\sum a_n\) is the harmonic series. 4. The series diverges.

The ratio test is inconclusive, and the series diverges because \(a_n=1/n\).

- Write the combinatorial quantity as products that can be compared across adjacent indices. - Cancel the common portions before simplifying the quotient. - What constant remains in the long run?

1. Let \(a_n=1/\binom{2n}{n}\). 2. Using the factorial form of the binomial coefficient, \(\frac{a_{n+1}}{a_n}=\frac{\binom{2n}{n}}{\binom{2n+2}{n+1}}=\frac{(n+1)^2}{(2n+2)(2n+1)}\). 3. Simplifying gives \(a_{n+1}/a_n=(n+1)/[2(2n+1)]\to1/4\). 4. Since \(1/4<1\), the series converges by the ratio test.

The series converges by the ratio test; the ratio-test limit is \(\frac14\).

- Check whether every consecutive-term ratio is actually defined. - Separate the zero positions from the nonzero positions. - Identify the series formed by the remaining magnitudes.

1. Infinitely many terms are \(0\), so ratios such as \(a_{2n+1}/a_{2n}\) are undefined. The usual consecutive-term ratio test is therefore not directly available. 2. The nonzero terms form the geometric subseries \(\sum_{n=1}^{\infty}a_{2n-1}=\sum_{n=1}^{\infty}3^{-n}\). 3. The even-indexed terms contribute \(0\). 4. Hence \(\sum|a_n|=\sum_{n=1}^{\infty}3^{-n}=\frac12\), so the series converges absolutely.

The consecutive-term ratio test is not directly applicable because infinitely many denominators in the ratios are \(0\). The series converges absolutely, with absolute sum \(\frac12\).

- Form the consecutive quotient before simplifying the variable powers. - Separate a bounded exponential-type factor from a factor that approaches zero. - Compare the product limit with the convergence boundary.

1. Let \(a_n=2^n/n^n\). 2. The consecutive-term ratio is \(a_{n+1}/a_n=\frac{2}{n+1}(n/(n+1))^n\). 3. The powered factor tends to \(e^{-1}\), while \(2/(n+1)\to0\), so the ratio tends to \(0\). 4. Therefore the series converges by the ratio test.

The series converges.

- What single factor remains after two adjacent finite products are divided? - How large can the oscillating correction be at late indices? - What constant does the remaining factor approach?

1. Consecutive products differ only by their final factor, so \(\left|\frac{a_{n+1}}{a_n}\right|=\frac{2+\frac{\sin(n+1)}{n+1}}{5}\). 2. Since \(\left|\sin(n+1)\right|\le1\), the fraction \(\sin(n+1)/(n+1)\) tends to \(0\). 3. The ratio-test limit is therefore \(2/5\). 4. Since \(2/5<1\), the series converges by the ratio test.

The series converges by the ratio test; the ratio-test limit is \(\frac25\).

- Separate the parameter factor from the rational index factor. - Analyze values away from the boundary first. - Inspect the original terms directly at the boundary values.

1. The absolute consecutive-term ratio is \(|x|/3\cdot(n/(n+1))^2\to|x|/3\). 2. The ratio test gives absolute convergence for \(|x|<3\) and divergence for \(|x|>3\). 3. If \(x=3\), the series is \(\sum1/n^2\). If \(x=-3\), its series of absolute values is also \(\sum1/n^2\). 4. Both boundary series converge absolutely.

The series converges absolutely for \(|x|\le3\) and diverges for \(|x|>3\).

- Expand \((2n+2)!\) by two factors and \((n+1)!\) by one factor. - Account for the extra factor of \(5\) in \(5^{n+1}\). - Compare the leading coefficients after all factorials cancel.

1. For \(a_n=\frac{(2n)!}{(n!)^2 5^n}\), \(\left|\frac{a_{n+1}}{a_n}\right|=\frac{(2n+2)(2n+1)}{5(n+1)^2}\). 2. Comparing leading coefficients gives \(L=\frac45<1\). 3. Therefore the series converges absolutely by the ratio test.

\(L=\frac45\), so the series converges absolutely by the ratio test.

- Expand the two new numerator factorial factors and the squared denominator factorial factor. - Include the extra factor of \(3\) from \(3^{n+1}\). - Decide domination by comparing the simplified ratio limit with \(1\).

1. For \(a_n=\frac{(2n)!}{(n!)^2 3^n}\), \(\left|\frac{a_{n+1}}{a_n}\right|=\frac{(2n+2)(2n+1)}{3(n+1)^2}\). 2. Comparing leading coefficients gives \(L=\frac43>1\). 3. Therefore the series diverges by the ratio test; the factor \(3^n\) does not dominate the central-factorial growth.

\(L=\frac43\), so the series diverges by the ratio test.

- Cancel \(n!\) after writing \((n+1)!=(n+1)n!\). - Rewrite the remaining quotient as the reciprocal of \(\left(1+\frac1n\right)^n\). - Use the standard limit defining \(e\).

1. For \(a_n=\frac{n!}{n^n}\), \(\left|\frac{a_{n+1}}{a_n}\right|=\frac{(n+1)!}{(n+1)^{n+1}}\frac{n^n}{n!}=\left(\frac{n}{n+1}\right)^n\). 2. Since \(\left(1+\frac1n\right)^n\to e\), the reciprocal limit is \(L=\frac1e<1\). 3. Therefore the series converges absolutely by the ratio test.

\(L=\frac1e\), so the series converges absolutely by the ratio test.

- Cancel the factor \(n+1\) from \((n+1)!\). - Express the remaining power quotient using \(1+\frac1n\). - Recognize the standard limit that approaches \(e\).

1. For \(a_n=\frac{n^n}{n!}\), \(\left|\frac{a_{n+1}}{a_n}\right|=\frac{(n+1)^{n+1}}{(n+1)!}\frac{n!}{n^n}=\left(1+\frac1n\right)^n\). 2. Thus \(L=e>1\). 3. Therefore the series diverges by the ratio test.

\(L=e>1\), so the series diverges by the ratio test.

- Expand both copies of \((n+1)!\) before canceling \((n!)^2\). - Combine the remaining variable powers into a single power of \(\frac{n}{n+1}\). - Relate that power to the standard exponential limit.

1. For \(a_n=\frac{(n!)^2}{n^{2n}}\), \(\left|\frac{a_{n+1}}{a_n}\right|=\frac{((n+1)!)^2}{(n+1)^{2n+2}}\frac{n^{2n}}{(n!)^2}=\left(\frac{n}{n+1}\right)^{2n}\). 2. Using \(\left(1+\frac1n\right)^n\to e\) gives \(L=\frac1{e^2}<1\). 3. Therefore the series converges absolutely by the ratio test.

\(L=\frac1{e^2}\), so the series converges absolutely by the ratio test.

- Expand both copies of \((n+1)!\) in the consecutive-term quotient. - Combine the remaining variable powers into \(\left(1+\frac1n\right)^{2n}\). - Apply the standard exponential limit and compare with \(1\).

1. For \(a_n=\frac{n^{2n}}{(n!)^2}\), \(\left|\frac{a_{n+1}}{a_n}\right|=\frac{(n+1)^{2n+2}}{((n+1)!)^2}\frac{(n!)^2}{n^{2n}}=\left(1+\frac1n\right)^{2n}\). 2. Using \(\left(1+\frac1n\right)^n\to e\) gives \(L=e^2>1\). 3. Therefore the series diverges by the ratio test.

\(L=e^2\), so the series diverges by the ratio test.

- Identify only the new odd and even factors added at index \(n+1\). - Include the additional factor from \(\left(\frac12\right)^{n+1}\). - Take the limit after canceling the shared product factors.

1. Passing from \(n\) to \(n+1\) adds the numerator factor \(2n+1\), the denominator factor \(2n+2\), and one factor of \(\frac12\). 2. Thus \(\left|\frac{a_{n+1}}{a_n}\right|=\frac{2n+1}{2n+2}\cdot\frac12\to\frac12\). 3. Since \(L=\frac12<1\), the series converges absolutely by the ratio test.

\(L=\frac12\), so the series converges absolutely by the ratio test.

- Cancel the product factors already present in both consecutive terms. - Keep only the new odd numerator factor and new even denominator factor. - Recall the ratio test’s conclusion at the boundary value \(L=1\).

1. Passing from \(n\) to \(n+1\) adds the factor \(2n+1\) to the numerator and \(2n+2\) to the denominator. 2. Therefore \(\left|\frac{a_{n+1}}{a_n}\right|=\frac{2n+1}{2n+2}\to1\). 3. The ratio-test limit is \(L=1\), so the ratio test is inconclusive.

\(L=1\), so the ratio test does not settle convergence.

- Expand the new factorial factors exactly as in a central-binomial quotient. - Include the extra factor of \(4\) from \(4^{n+1}\). - Treat the boundary value \(L=1\) separately from the decisive ratio-test cases.

1. For \(a_n=\frac{(2n)!}{(n!)^2 4^n}\), \(\left|\frac{a_{n+1}}{a_n}\right|=\frac{(2n+2)(2n+1)}{4(n+1)^2}\). 2. Comparing leading coefficients gives \(L=1\). 3. Therefore the ratio test is inconclusive.

\(L=1\), so the ratio test is inconclusive.

- Handle \(c=0\) directly before canceling powers of \(c\). - For \(c\ne0\), simplify the central-factorial quotient and include the factor \(|c|\). - Compare the resulting limit with \(1\), keeping the equality case separate.

1. If \(c=0\), every term is \(0\), so the series converges absolutely. 2. If \(c\ne0\), then for \(a_n=\frac{(2n)!}{(n!)^2}c^n\), \(\left|\frac{a_{n+1}}{a_n}\right|=|c|\frac{(2n+2)(2n+1)}{(n+1)^2}\to4|c|\). 3. The ratio test proves absolute convergence when \(0<|c|<\frac14\) and divergence when \(|c|>\frac14\). 4. When \(|c|=\frac14\), the ratio test is inconclusive.

The series converges absolutely for \(|c|<\frac14\) (directly when \(c=0\), and by the ratio test when \(0<|c|<\frac14\)), diverges for \(|c|>\frac14\), and the ratio test is inconclusive for \(|c|=\frac14\).

- Form the consecutive ratio after including the polynomial weight. - Determine the limit of the new factor introduced by that weight. - Compare the resulting ratio limit with the original one.

1. Let \(b_n=n^k a_n\). 2. Its absolute consecutive-term ratio is \(\left|\frac{b_{n+1}}{b_n}\right|=\left(\frac{n+1}{n}\right)^k\left|\frac{a_{n+1}}{a_n}\right|.\) 3. The polynomial factor approaches \(1\), so the ratio limit for \((b_n)\) is \(1\cdot L=L<1\). 4. The ratio test therefore proves that \(\sum|n^k a_n|\) converges.

For every fixed nonnegative integer \(k\), \(\sum n^k a_n\) converges absolutely.

- Form the quotient of consecutive terms before expanding more than necessary. - Separate the ordinary rational factor from the large repeated-power factor. - What standard limiting behavior appears in that repeated-power factor?

1. Let \(a_n=n!/(n+2)^n\). 2. The consecutive-term ratio is \(\left|\frac{a_{n+1}}{a_n}\right|=\frac{n+1}{n+3}\left(\frac{n+2}{n+3}\right)^n\). 3. The first factor tends to \(1\), and the variable-power factor tends to \(e^{-1}\). Thus the ratio-test limit is \(1/e\). 4. Since \(1/e<1\), the series converges by the ratio test.

The series converges by the ratio test; the ratio-test limit is \(\frac1e\).

- Expand only the new factors in the large denominator product. - Rewrite the shifted numerator power relative to the previous one. - Which remaining factor grows without bound?

1. Let \(a_n=n^n/(2n)!\). 2. The consecutive-term ratio is \(\left|\frac{a_{n+1}}{a_n}\right|=\frac{(n+1)^{n+1}}{n^n(2n+1)(2n+2)}=\frac{(1+1/n)^n}{2(2n+1)}\). 3. The numerator tends to \(e\) while the denominator grows without bound, so the ratio tends to \(0\). 4. Therefore the series converges by the ratio test.

The series converges by the ratio test; the ratio-test limit is \(0\).

- Make one example with an exact repeated multiplicative pattern. - In the second example, preserve the same eventual change while altering the earlier or variable behavior. - Compute the complete totals rather than comparing only limiting changes.

1. Let \(a_n=(4/5)^{n-1}\). Then \(a_1=1\), every ratio equals \(4/5\), and \(\sum a_n=5\). 2. Define \(b_1=1\) and \(b_n=2(4/5)^{n-1}\) for \(n\ge2\). 3. For \(n\ge2\), \(b_{n+1}/b_n=4/5\), so the ratio limit is also \(0.8\), and \(b_1=1\). 4. However, \(\sum b_n=1+2\sum_{n=2}^{\infty}(4/5)^{n-1}=1+2(4)=9\). A limiting ratio determines convergence behavior, not the sum.

The claim is false. The two examples have sums \(5\) and \(9\) despite sharing first term \(1\) and ratio limit \(0.8\).

- The recurrence directly gives the quotient of consecutive magnitudes. - Which parameter values place its limiting size below or above the boundary? - At the boundary, inspect whether the magnitudes can approach zero.

1. The absolute consecutive-term ratio is \(\left|\frac{a_{n+1}}{a_n}\right|=|c|\frac{n+2}{3n+1}\to\frac{|c|}{3}\). 2. The ratio test gives absolute convergence for \(|c|<3\) and divergence for \(|c|>3\). 3. If \(|c|=3\), then \(\left|a_{n+1}/a_n\right|=(3n+6)/(3n+1)>1\). 4. Thus \(|a_n|\) does not approach \(0\) at the boundary, so the series diverges there by the \(n\)th-term test.

The series converges absolutely for \(|c|<3\) and diverges for \(|c|\ge3\).

- First use the recurrence itself to obtain the ratio-test threshold. - At equality, solve the recurrence explicitly rather than relying on the inconclusive test. - The two boundary signs produce different familiar series.

1. The absolute ratio tends to \(|c|/2\). Thus the ratio test gives absolute convergence for \(|c|<2\) and divergence for \(|c|>2\). 2. The recurrence telescopes to \(a_n=(c/2)^{n-1}/n\). 3. If \(c=2\), then \(a_n=1/n\), so the series diverges. 4. If \(c=-2\), then \(a_n=(-1)^{n-1}/n\), so the series converges conditionally.

The series converges absolutely for \(|c|<2\), converges conditionally for \(c=-2\), and diverges for \(c=2\) or \(|c|>2\).

- Combine two consecutive ratio steps instead of examining only one. - Separate the terms according to parity. - Recognize the resulting constant-ratio subseries.

1. Over any two consecutive steps, \(\left|a_{n+2}/a_n\right|=(1/2)(3/4)=3/8\), regardless of the parity of the starting index. 2. Thus the odd-indexed magnitudes form a geometric sequence with ratio \(3/8\), and the even-indexed magnitudes form another geometric sequence with ratio \(3/8\). 3. Both \(\sum|a_{2k-1}|\) and \(\sum|a_{2k}|\) converge. 4. Their sum is \(\sum|a_n|\), so the original series converges absolutely.

The series converges absolutely, even though the one-step ratio limit does not exist.

- First determine the limiting size of an individual term. - Then examine what happens to the quotient of consecutive terms. - When that quotient reaches the boundary, which necessary condition settles the question?

1. The term can be written as \(a_n=\left(1+\frac{c-1}{n+1}\right)^n\), so \(a_n\to e^{c-1}>0\). 2. Since both \(a_{n+1}\) and \(a_n\) approach the same positive limit, \(\left|a_{n+1}/a_n\right|\to1\). 3. The ratio test is inconclusive because its limit is \(1\). 4. The terms do not approach \(0\), so the series diverges by the \(n\)th-term test.

The ratio-test limit is \(1\), so that test is inconclusive. The series diverges because its terms approach \(e^{c-1}\ne0\).

- First use the recurrence exactly as written to examine consecutive terms. - An inconclusive test does not settle the series; unfold several recurrence steps and look for cancellation. - Rewrite the resulting general term in a form whose partial sums simplify.

1. The recurrence gives \(\frac{a_{n+1}}{a_n}=\frac{n}{n+2}\to1\), so the ratio test is inconclusive. 2. Expanding the recurrence, \(a_n=\prod_{k=1}^{n-1}\frac{k}{k+2} =\frac{(n-1)!}{3\cdot4\cdots(n+1)} =\frac{2}{n(n+1)}\). 3. Rewrite \(a_n=2\left(\frac1n-\frac{1}{n+1}\right)\). 4. The partial sums telescope: \(\sum_{n=1}^{N}a_n=2\left(1-\frac{1}{N+1}\right)\to2\).

a) The ratio-test limit is \(1\), so the test is inconclusive. b) The series converges, and its sum is \(2\).

- Distinguish a limiting consecutive ratio from an exact common ratio. - Use the limiting change only to establish convergence. - To find the total, compare the series with a shifted scalar multiple of itself.

1. For \(a_n=n/2^n\), \(a_{n+1}/a_n=(n+1)/(2n)\to1/2\), so the ratio test proves convergence. 2. A ratio limit of \(1/2\) does not mean every consecutive ratio equals \(1/2\); the series is not geometric. 3. Let \(S=\sum_{n=1}^{\infty}n/2^n\). Then \(S/2=\sum_{n=1}^{\infty}n/2^{n+1}\). 4. Subtracting aligned series gives \(S/2=1/2+1/4+1/8+\cdots=1\), so \(S=2\).

The ratio test proves convergence, but the series is not geometric. Its actual sum is \(2\).

- How many new product factors appear when the squared index advances? - Can every one of those factors be bounded below by the same simple quantity? - Use that bound to control the entire quotient.

1. Let \(a_n=(n!)^2/(n^2)!\). 2. The consecutive-term ratio is \(\frac{a_{n+1}}{a_n}=\frac{(n+1)^2}{(n^2+1)(n^2+2)\cdots(n^2+2n+1)}\). 3. Every denominator factor is at least \(n^2+1\), so \(0\le\frac{a_{n+1}}{a_n}\le\frac{(n+1)^2}{(n^2+1)^{2n+1}}\to0\). 4. Therefore the ratio-test limit is \(0\), and the series converges.

The series converges by the ratio test; the ratio-test limit is \(0\).

- Cancel the shared product factors to obtain the ratio-test limit. - If the test is inconclusive, simplify the finite product itself. - Check the necessary term condition after telescoping.

1. The ratio is \(a_{n+1}/a_n=1-1/(n+1)^2\to1\), so the ratio test is inconclusive. 2. Factor each product term as \((k-1)(k+1)/k^2\). 3. The product telescopes to \(a_n=(n+1)/(2n)\). 4. Since \(a_n\to1/2\ne0\), the series diverges by the \(n\)th-term test.

The ratio test is inconclusive, and the series diverges because \(a_n\to\frac12\).

- Keep both denominator terms while forming the consecutive quotient. - Normalize by the product term common to adjacent indices. - Which grows faster in the long run: the product or the fixed-base exponential?

1. Let \(a_n=1/(n!+3^n)\). Then \(\frac{a_{n+1}}{a_n}=\frac{n!+3^n}{(n+1)!+3^{n+1}}\). 2. Dividing numerator and denominator by \(n!\) gives \(\frac{1+3^n/n!}{n+1+3(3^n/n!)}\). 3. The auxiliary sequence \(3^n/n!\) tends to \(0\) because its consecutive-term ratio is \(3/(n+1)\to0\). 4. The numerator therefore tends to \(1\), while the denominator grows without bound, so the ratio-test limit is \(0\). 5. Hence the series converges.

The series converges by the ratio test; the ratio-test limit is \(0\).

- Preserve both square exponents when adjacent terms are divided. - Regroup the quotient into factors whose bases approach one. - Evaluate the logarithmic size of each factor before combining them.

1. Let \(a_n=(n/(n+2))^{n^2}\). Then \(\frac{a_{n+1}}{a_n}=\left(\frac{n+1}{n+3}\right)^{(n+1)^2}\left(\frac{n+2}{n}\right)^{n^2}\). 2. Regrouping gives \(\frac{a_{n+1}}{a_n}=\left(1+\frac{2}{n^2+3n}\right)^{n^2}\left(1-\frac{2}{n+3}\right)^{2n+1}\). 3. The first factor tends to \(e^2\), and the second tends to \(e^{-4}\). 4. The ratio-test limit is therefore \(e^{-2}\). 5. Since \(e^{-2}<1\), the series converges.

The series converges by the ratio test; the ratio-test limit is \(e^{-2}\).

- Separate the parameter from the index-dependent portion of the consecutive quotient. - How close is the powered base to one at large indices? - At any boundary value, check the original term limit directly.

1. If \(c=0\), every term is \(0\), so the series converges absolutely. 2. For \(c\ne0\), let \(a_n=(c(n+1)/(5n+2))^n\). Then \(\left|\frac{a_{n+1}}{a_n}\right|=|c|\frac{n+2}{5n+7}\left(\frac{(n+2)(5n+2)}{(5n+7)(n+1)}\right)^n\). 3. The powered base equals \(1-\frac{3}{5n^2+12n+7}\), so its \(n\)th power tends to \(1\). 4. The ratio-test limit is \(|c|/5\). Thus the series converges absolutely for \(|c|<5\) and diverges for \(|c|>5\). 5. When \(|c|=5\), \(|a_n|=(1+3/(5n+2))^n\to e^{3/5}\ne0\), so the series diverges by the \(n\)th-term test.

The series converges absolutely for \(|c|<5\) and diverges for \(|c|\ge5\).