- Set the absolute value of the common ratio less than \(1\). - The radius is the distance from the center \(0\) to either boundary.

1. The common ratio is \(\frac{x}{3}\). 2. A geometric series converges when \(\left|\frac{x}{3}\right|<1\), which gives \(|x|<3\). Therefore, the radius is \(R=3\).

\(R=3\).

- Compare the condition with the standard form \(|x-c|<R\). - Rewrite \(x+2\) as \(x-(-2)\).

1. Write the condition as \(|x-(-2)|<4\). 2. Therefore, the center is \(c=-2\) and the radius is \(R=4\).

\(c=-2\) and \(R=4\).

- Measure each input's distance from the center \(5\). - Compare that distance with the radius \(2\). - A boundary point has distance exactly equal to the radius; its convergence requires a separate endpoint test.

1. Compare each distance from the center \(5\) with the radius \(2\). 2. For \(x=6\), \(|6-5|=1<2\), so the point is inside the radius. 3. For \(x=3\), \(|3-5|=2\), so the point is on the boundary. 4. For \(x=2\), \(|2-5|=3>2\), so the point is outside the radius.

a) Inside the radius. b) On the boundary. c) Outside the radius.

- Calculate the distance from \(x=1\) to the center. - Recall what a radius tells you about boundary points.

1. The point \(x=1\) is exactly \(3\) units from the center, so it is on the boundary. 2. The radius alone does not determine endpoint behavior. The series may converge or diverge at this point.

Either can happen. \(x=1\) is a boundary point, and the radius does not settle convergence there.

- Read the center from the expression \(x-2\). - Convert the geometric convergence condition into a distance inequality. - At an endpoint, inspect the actual common ratio.

1. The series is geometric with common ratio \(\frac{x-2}{4}\). 2. It converges when \(\left|\frac{x-2}{4}\right|<1\), or \(|x-2|<4\). Thus the radius is \(R=4\). 3. The inequality gives the preliminary interval \((-2, 6)\). 4. At \(x=-2\), the common ratio is \(-1\), so the terms do not approach \(0\) and the series diverges. 5. At \(x=6\), the common ratio is \(1\), so the terms again fail to approach \(0\). The interval is \((-2, 6)\).

Radius: \(R=4\). Interval of convergence: \((-2, 6)\). Both endpoints diverge.

- Check the expansion center separately, where terms may all vanish. - Keep the exponent \(2n\) intact when forming the ratio at other inputs. - Test the endpoints in the original series after finding the radius.

1. At \(x=2\), every term is zero, so the series converges. For \(x\neq2\), let \(a_n=\frac{(x-2)^{2n}}{n\,9^n}\). 2. The ratio is \(\left|\frac{a_{n+1}}{a_n}\right|=\frac{|x-2|^2}{9}\frac{n}{n+1}\to\frac{|x-2|^2}{9}\). 3. The ratio test guarantees convergence for \(|x-2|<3\), giving radius \(3\). 4. At both \(x=-1\) and \(x=5\), the series is \(\sum_{n=1}^{\infty}\frac{1}{n}\), which diverges. The interval is \((-1,5)\).

Radius \(3\); interval \((-1,5)\).

- Consider the center separately if consecutive terms both equal zero. - Simplify the factorial quotient in the ratio. - Ask whether the resulting limit depends on the chosen fixed input.

1. At \(x=-1\), all terms vanish and the series converges. For \(x\neq-1\), write \(u_n=\frac{n^2(x+1)^n}{n!}\). 2. The ratio of consecutive absolute values is \(\left|\frac{u_{n+1}}{u_n}\right|=\frac{n+1}{n^2}|x+1|\). 3. For every fixed \(x\neq-1\), the limit is \(0\), so the series converges absolutely. Including the center, it converges for all real \(x\).

Radius \(R=\infty\); interval \((-\infty,\infty)\).

- Compare consecutive terms before trying to solve an interval inequality. - For a nonzero distance from the center, the factorial growth dominates. - Check the center separately after ruling out other values.

1. Let \(u_n=n!(x-4)^n\). For \(x\neq4\), \( \left|\frac{u_{n+1}}{u_n}\right|=(n+1)|x-4|\). 2. This ratio grows without bound, so the terms cannot approach \(0\). The series diverges for every \(x\neq4\). 3. At \(x=4\), every term is \(0\), so the series converges. 4. Therefore, the radius is \(R=0\), and the set of convergence is the single point \(\{4\}\).

Radius: \(R=0\). Set of convergence: \(\{4\}\).

- Keep the factor \(\frac{n+1}{n}\) until taking the limit. - Convert the ratio-test result into a distance from the center. - At an endpoint, first check whether the terms approach \(0\).

1. For the ratio of consecutive absolute values, \( \left|\frac{u_{n+1}}{u_n}\right|=\frac{n+1}{n}\cdot\frac{|x-4|}{3}\). 2. The limit is \(\frac{|x-4|}{3}\), so the series converges when \(|x-4|<3\). Thus \(R=3\) and the preliminary interval is \((1, 7)\). 3. At \(x=1\), the terms are \(n(-1)^n\), which do not approach \(0\). 4. At \(x=7\), the terms are \(n\), which also do not approach \(0\). 5. Therefore, the interval of convergence is \((1, 7)\).

Radius: \(R=3\). Interval: \((1, 7)\). Both endpoints diverge.

- Treat the entire expression multiplying itself with each index as the geometric ratio. - Solve the convergence inequality as a squared-distance inequality. - Substitute either endpoint into the ratio to test convergence.

1. Rewrite the series as \( \sum_{n=0}^{\infty}\left[9(x-2)^2\right]^n\). 2. This geometric series converges when \( |9(x-2)^2|<1\). 3. Since the expression is nonnegative, the inequality becomes \((x-2)^2<\frac{1}{9}\), or \(|x-2|<\frac{1}{3}\). 4. Thus the radius is \(R=\frac{1}{3}\), with preliminary interval \(\left(\frac{5}{3}, \frac{7}{3}\right)\). 5. At either endpoint, the geometric ratio equals \(1\), so the terms do not approach \(0\). Both endpoints diverge.

Radius: \( R=\frac13\). Interval: \( \left(\frac53, \frac73\right)\).

- The center lies halfway between the two boundary values. - The radius is half the total length of the interval. - Read endpoint inclusion directly from the bracket and parenthesis notation.

1. The center is the midpoint of the interval endpoints: \( c=\frac{-4+2}{2}=-1\). 2. The radius is the distance from the center to either endpoint: \( R=2-(-1)=3\). 3. The bracket at \(-4\) means the series converges there. The parenthesis at \(2\) means it diverges there.

a) Center \(c=-1\). b) Radius \(R=3\). c) The series converges at \(x=-4\) and diverges at \(x=2\).

- Distribute the given factor into each term. - Identify the part of the term that depends exponentially on the index. - Check the two endpoints and the center separately.

1. Multiplying term by term gives \(G(x)=\sum_{n=0}^{\infty}\frac{(x-2)^{n+4}}{3^n}\). 2. For \(x\neq2\), the factor \((x-2)^4\) is finite and nonzero, so this series converges exactly where the underlying geometric series \(F\) does. At \(x=2\), both converge because all terms of the product vanish. 3. The geometric condition \(\left|\frac{x-2}{3}\right|<1\) gives radius \(3\). At \(x=-1\) and \(x=5\), terms fail to approach zero, so the interval is \((-1,5)\). 4. In this particular multiplication by a power of \(x-2\), only the starting exponent changes; the geometric factor governing successive terms remains \(\frac{x-2}{3}\).

a) \(G(x)=\sum_{n=0}^{\infty}\frac{(x-2)^{n+4}}{3^n}\). b) Radius \(3\); interval \((-1,5)\). c) The same geometric ratio determines convergence away from the center, which also converges.

- Replace the original series variable by \(2x-1\) in the radius condition. - Solve the resulting absolute-value inequality before identifying the midpoint. - A radius determines the open interval but never decides boundary behavior by itself.

1. The original series converges when \(|u|<4\). 2. For \(G\), the input is \(u=2x-1\), so convergence requires \( |2x-1|<4\). 3. Solving the inequality gives \( -\frac32<x<\frac52\). 4. The midpoint of this interval is \(\frac12\), and the distance from the midpoint to either endpoint is \(2\). Thus the center is \(\frac12\) and the radius is \(2\). 5. At the endpoints, the original input equals \(-4\) or \(4\). A radius alone does not determine convergence at those boundary values, so both endpoints must remain undecided without more information about the coefficients.

a) Center \( \frac12\); radius \(2\). b) \( \left(-\frac32, \frac52\right)\). c) Endpoint convergence cannot be determined from the radius alone.

- Compare each point's distance from \(-2\) with the radius \(5\). - Points strictly inside the radius converge, and points strictly outside diverge. - Boundary points always require separate information about the coefficients.

1. The series converges whenever the distance from the center is less than the radius: \( |x+2|<5\). This gives the open interval \((-7, 3)\). 2. For \(x=1\), the distance from the center is \(3<5\), so the series definitely converges. 3. For \(x=-8\), the distance is \(6>5\), so the series definitely diverges. 4. The points \(x=3\) and \(x=-7\) are exactly \(5\) units from the center. Radius information alone does not determine convergence at endpoints.

a) \((-7, 3)\). b) \(x=1\): definitely convergent; \(x=-8\): definitely divergent; \(x=3\): undetermined; \(x=-7\): undetermined.

- Recall exactly what the ratio test says when its limit equals \(1\). - Use the ratio result only to find the open interval first. - Substitute each endpoint into the original series and choose a new convergence test.

1. A ratio-test limit of \(1\) is inconclusive. It proves neither convergence nor divergence, so each endpoint requires a different test. 2. The ratio test gives convergence when \(\frac{|x-2|}{3}<1\), so \(|x-2|<3\). The center is \(2\), the radius is \(3\), and the preliminary interval is \((-1,5)\). 3. At \(x=5\), the series becomes \(\sum_{n=1}^{\infty}\frac{1}{n^2}\), which converges absolutely. 4. At \(x=-1\), the series becomes \(\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}\). Its absolute-value series is also \(\sum_{n=1}^{\infty}\frac{1}{n^2}\), so it converges absolutely. 5. Therefore, both endpoints are included and the interval of convergence is \([-1,5]\).

a) A ratio-test limit of \(1\) is inconclusive. b) Radius \(R=3\); interval \([-1,5]\). c) The series converges absolutely at both endpoints.

- Compare distances from each point to the center. - Convergence at one noncenter point forces absolute convergence at every strictly closer point. - Endpoint inclusion may vary, but the interior of the radius cannot contain gaps.

1. The point \(7\) is \(5\) units from the center \(2\). Convergence at that point guarantees absolute convergence at every point closer to the center. 2. Therefore, the series must converge absolutely whenever \(|x-2|<5\), which is the interval \((-3,7)\). The claimed set excludes points such as \(x=6\), so it is impossible. 3. For a finite positive radius \(R\), the real convergence set is one of \((c-R,c+R)\), \([c-R,c+R)\), \((c-R,c+R]\), or \([c-R,c+R]\). If \(R=0\), only the center is guaranteed. If \(R=\infty\), the series converges for every real \(x\). 4. Thus convergence at \(x=7\) guarantees absolute convergence throughout \((-3,7)\). It does not determine what happens at the opposite boundary point \(-3\).

a) Convergence at \(7\) forces absolute convergence throughout \((-3,7)\), so the claimed set cannot have a gap. b) For finite \(R>0\), the open radius interval has neither, either, or both endpoints included; \(R=0\) gives only the center, and \(R=\infty\) gives all real numbers. c) Absolute convergence is guaranteed for every \(x\in(-3,7)\).

- Measure the horizontal distance from the center marker to either boundary marker. - Recall what a radius of convergence guarantees strictly inside and strictly outside the radius. - Boundary points satisfy \(|x-c|=R\), so ask whether the general theorem settles equality cases.

1. The number line places the center at \(C=1\) and the boundary points at \(A=-2\) and \(B=4\). Each boundary is \(3\) units from the center, so \(c=1\) and \(R=3\). 2. A power series converges absolutely when \(|x-c|<R\), so it is guaranteed to converge absolutely for \(-2<x<4\). It diverges when \(|x-c|>R\), so it is guaranteed to diverge for \(x<-2\) or \(x>4\). 3. At \(x=-2\) and \(x=4\), the distance from the center equals the radius. General radius-of-convergence information does not determine endpoint behavior, so each endpoint must be tested separately.

a) \(c=1\) and \(R=3\). b) Guaranteed absolute convergence on \((-2,4)\); guaranteed divergence for \(x<-2\) or \(x>4\). c) Convergence at \(x=-2\) and \(x=4\) is undetermined without separate endpoint tests.

- First rewrite the ratio in terms of the distance from its center. - Compare the resulting convergence inequality with the form \(|x-c|<R\). - After choosing the parameter, check the two boundary values separately.

1. Rewrite the geometric ratio as \(\frac{2x-6}{a}=\frac{2(x-3)}{a}\). 2. A geometric series converges when \(\left|\frac{2(x-3)}{a}\right|<1\), which is equivalent to \(|x-3|<\frac{a}{2}\). Therefore the radius is \(\frac{a}{2}\). 3. Requiring radius \(5\) gives \(\frac{a}{2}=5\), so \(a=10\). 4. Then \(|x-3|<5\), giving \(-2<x<8\). At either endpoint the geometric ratio is \(1\) or \(-1\), so the terms do not approach \(0\). The interval of convergence is \((-2, 8)\).

\(a=10\), and the interval of convergence is \((-2, 8)\).

- In the ratio, cancel common powers before taking the limit. - Express the condition as a distance from the center.

1. At \(x=1\), every term is zero. Elsewhere, the ratio of consecutive absolute terms is \(\frac{n+1}{n}\frac{|x-1|}{5}\). 2. The limit is \(\frac{|x-1|}{5}\); the series converges when this is less than 1. 3. Thus the radius is \(R=5\).

\(R=5\).

- Check whether the factorial is in the numerator or denominator. - Form a consecutive-term ratio for each series. - Consider the center separately when all higher powers vanish.

1. For A, the ratio of consecutive absolute terms is \((n+1)|x|\), which grows without bound for any \(x\neq0\). At \(x=0\), the series converges, so \(R_A=0\). 2. For B, the ratio is \(\frac{|x|}{n+1}\to0\) for every real \(x\). Thus \(R_B=\infty\). 3. Factorials multiplying terms in A make them grow for nonzero inputs; factorials in the denominator of B make terms shrink quickly.

\(R_A=0\) and \(R_B=\infty\).

- A geometric series needs a ratio whose absolute value is less than 1. - First find the input values where the geometric ratio has magnitude \(1\). - Check the behavior at the boundary before choosing the marked readings.

1. The series is geometric with ratio \(\frac{v-5}{2}\). 2. It converges when \(\left|\frac{v-5}{2}\right|<1\), equivalent to \(3<v<7\). 3. Markers B at \(4\) and C at \(6\) are inside that interval. A at \(3\) and D at \(7\) are endpoints where the geometric series diverges.

B and C. The convergence interval is \((3,7)\,\text{V}\).

- First ignore the endpoints and use the ratio test to find the radius. - Substitute each endpoint into the original power series. - Distinguish ordinary convergence from absolute convergence at the alternating endpoint.

1. The ratio of consecutive absolute values approaches \(\frac{|x-1|}{2}\). Therefore, the series converges absolutely when \(|x-1|<2\), giving radius \(R=2\) and preliminary interval \((-1, 3)\). 2. At \(x=3\), the series becomes \( \sum_{n=1}^{\infty}\frac{1}{n}\), which diverges. 3. At \(x=-1\), the series becomes \( \sum_{n=1}^{\infty}\frac{(-1)^n}{n}\), which converges conditionally. 4. Thus the interval of convergence is \([-1, 3)\).

Radius: \(R=2\). Interval: \([-1, 3)\). At \(x=-1\): conditional convergence. At \(x=3\): divergence.

- Test the original endpoints with a \(p\)-series comparison. - Differentiate the coefficient and power together. - Recheck both endpoints after differentiation; they are not inherited automatically.

1. The radius is \(1\). At \(x=1\), the series is \( \sum_{n=1}^{\infty}\frac{1}{n^2}\), and at \(x=-1\), its absolute-value series is also \( \sum_{n=1}^{\infty}\frac{1}{n^2}\). Thus \(F\) converges on \([-1, 1]\). 2. Differentiating gives \( F'(x)=\sum_{n=1}^{\infty}\frac{x^{n-1}}{n}\). 3. The differentiated series still has radius \(1\). 4. At \(x=1\), it becomes the harmonic series and diverges. At \(x=-1\), it becomes an alternating harmonic series and converges conditionally. 5. Therefore, the differentiated series has interval \([-1, 1)\).

a) Radius \(1\); interval \([-1, 1]\). b) \( F'(x)=\sum_{n=1}^{\infty}\frac{x^{n-1}}{n}\), with interval \([-1, 1)\). c) Differentiation preserved the radius but changed endpoint convergence.

- Generate the first few coefficients to recognize the pattern. - Substitute the closed formula into the power series. - Treat the resulting expression as a geometric series in the shifted variable.

1. The recurrence doubles each term, so \(a_n=3\cdot2^n\). 2. The power series becomes \( 3\sum_{n=0}^{\infty}[2(x+1)]^n\). 3. This geometric series converges when \(|2(x+1)|<1\), or \(|x+1|<\frac{1}{2}\). Thus \(R=\frac{1}{2}\). 4. The preliminary interval is \(\left(-\frac{3}{2}, -\frac{1}{2}\right)\). At either endpoint, the geometric ratio is \(1\) or \(-1\), so the terms do not approach \(0\).

a) \(a_n=3\cdot2^n\). b) Radius \( R=\frac12\); interval \( \left(-\frac32, -\frac12\right)\).

- Find each component's convergence behavior. - A sum may exhibit cancellation; check whether either component dominates outside the smaller radius. - Test the two boundary inputs separately in the original expression.

1. The first component is \(\sum_{n=1}^{\infty}\frac{1}{n}\left(\frac{(x+1)^2}{9}\right)^n\). It converges for \(|x+1|<3\) and diverges at \(|x+1|=3\), where it becomes the harmonic series. 2. The second component is \(\sum_{n=1}^{\infty}\frac{1}{n^2}\left(\frac{x+1}{4}\right)^n\). It converges absolutely for \(|x+1|\leq4\). 3. On \(|x+1|<3\), both components converge, so their sum converges. If \(|x+1|>3\), the positive first component grows in magnitude termwise and dominates the second: the ratio of their magnitudes is \(n\left(\frac{4|x+1|}{9}\right)^n\to\infty\). Thus the combined terms fail to approach zero; cancellation cannot rescue convergence. 4. At \(x=-4\) and \(x=2\), the first component is harmonic while the second converges absolutely. Their sum therefore diverges at either endpoint. 5. The center is \(-1\), the radius is \(3\), and the interval is \((-4,2)\).

Radius \(R=3\); interval \((-4,2)\). The first component sets the radius; the combined series diverges at both endpoints.

- First ignore the endpoints and determine the common open interval. - Substitute each endpoint before choosing a convergence test. - Compare how \(1\), \(1/n\), and \(1/n^2\) change the endpoint behavior.

1. For each fixed value of \(p\), the exponential factor \(4^n\) determines the radius. All three series have center \(2\), radius \(R=4\), and open interval \((-2, 6)\). 2. At \(x=6\), the series becomes \( \sum_{n=1}^{\infty}\frac{1}{n^p}\). It diverges for \(p=0\) and \(p=1\), and converges absolutely for \(p=2\). 3. At \(x=-2\), the series becomes \( \sum_{n=1}^{\infty}\frac{(-1)^n}{n^p}\). For \(p=0\), its terms do not approach \(0\), so it diverges. For \(p=1\), it converges conditionally by the alternating series test. For \(p=2\), it converges absolutely. 4. Therefore, the intervals are \((-2, 6)\) for \(p=0\), \([-2, 6)\) for \(p=1\), and \([-2, 6]\) for \(p=2\).

\(p=0\): radius \(4\), interval \((-2, 6)\). \(p=1\): radius \(4\), interval \([-2, 6)\), conditional at \(x=-2\). \(p=2\): radius \(4\), interval \([-2, 6]\), absolute at both endpoints.

- Group the centered power and the scale factor into one new variable. - Recognize the resulting odd-power alternating series. - At each endpoint, test both ordinary and absolute convergence.

1. Let \( u=\frac{x-2}{4}\). The series becomes \( \sum_{n=0}^{\infty}\frac{(-1)^nu^{2n+1}}{2n+1}\), which is the Maclaurin series for \(\arctan u\). 2. The arctangent series converges for \(|u|<1\), so \( \left|\frac{x-2}{4}\right|<1\). Thus the center is \(2\), the radius is \(4\), and the open interval is \((-2, 6)\). 3. At \(x=6\), the series is \( 1-\frac13+\frac15-\frac17+\cdots\), which converges conditionally. 4. At \(x=-2\), the series is the negative of that alternating series, so it also converges conditionally. 5. Therefore, the interval of convergence is \([-2, 6]\).

Radius: \(R=4\). Interval: \([-2, 6]\). Convergence is conditional at both endpoints.

- Use partial fractions so that each term can be rewritten as a geometric series. - For each center, write the convergence condition in terms of the displacement from that center. - The common radius is controlled by the more restrictive geometric series; then test both boundary points.

1. Partial fractions give \( f(x)=\frac17\left(\frac{1}{x+2}+\frac{1}{5-x}\right)\). 2. Centered at \(0\), the two geometric expansions require \(|x|<2\) and \(|x|<5\), respectively. Their common condition is \(|x|<2\), so \(R=2\). 3. At \(x=-2\), the function is undefined. At \(x=2\), the geometric series from \(\frac{1}{x+2}\) has ratio \(-1\), so its terms do not approach \(0\). Thus the interval is \((-2, 2)\). 4. For center \(2\), let \(h=x-2\). Then \( \frac{1}{x+2}=\frac14\frac{1}{1+\frac{h}{4}}\) requires \(|h|<4\), while \( \frac{1}{5-x}=\frac13\frac{1}{1-\frac{h}{3}}\) requires \(|h|<3\). Therefore, \(R=3\). 5. At \(h=3\) and \(h=-3\), the second geometric series has ratio \(1\) or \(-1\), so its terms do not approach \(0\). Since \(x=2+h\), the interval is \((-1, 5)\). 6. A Taylor series reaches only to the nearest point where the function is undefined. The nearest singularity is \(2\) units from center \(0\), but \(3\) units from center \(2\).

a) Center \(0\): \(R=2\), interval \((-2, 2)\). b) Center \(2\): \(R=3\), interval \((-1, 5)\). c) The distance from the chosen center to the nearest singularity changes.

- First use the ratio test to obtain the open interval. - Test the two endpoints in the original series, not in the ratio-test expression. - Compare what changes and what stays unchanged when only finitely many terms are altered.

1. The ratio test gives convergence when \( \frac{|x-4|}{3}<1\), so the radius is \(R=3\) and the open interval is \((1, 7)\). 2. At \(x=7\), the series becomes \( \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\), which converges conditionally. 3. At \(x=1\), the series becomes \( -\sum_{n=1}^{\infty}\frac1n\), which diverges. Therefore, \(F\) has interval \((1, 7]\). 4. Removing finitely many terms does not change whether an infinite series converges. Thus \(T\) also has radius \(3\) and interval \((1, 7]\). 5. A polynomial is a finite sum and is defined for every real \(x\). Adding it changes only finitely many coefficients, so it cannot change the convergence of the infinite tail. Therefore, \(T(x)+P(x)\) also has interval \((1, 7]\).

a) Radius \(R=3\); interval \((1, 7]\). b) Radius \(R=3\); interval \((1, 7]\). c) Adding or removing finitely many terms does not change convergence of the infinite tail, so the interval remains \((1, 7]\).

- Integrate one generic term before rewriting the index. - Test \(x=1\) and \(x=-1\) separately for each series. - Focus on the new denominator factors when comparing endpoint behavior.

1. Integrating term by term gives \( F_1(x)=\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}=\sum_{m=1}^{\infty}\frac{x^m}{m}\). 2. Integrating again gives \( F_2(x)=\sum_{m=1}^{\infty}\frac{x^{m+1}}{m(m+1)}=\sum_{j=2}^{\infty}\frac{x^j}{(j-1)j}\). 3. For \(F_0\), the terms fail to approach \(0\) at both \(x=1\) and \(x=-1\), so its interval is \((-1, 1)\). 4. For \(F_1\), the series at \(x=1\) is harmonic and diverges. At \(x=-1\), it is alternating harmonic and converges. Thus its interval is \([-1, 1)\). 5. For \(F_2\), the absolute-value series at either endpoint is \( \sum_{j=2}^{\infty}\frac{1}{(j-1)j}\), which telescopes and converges. Thus its interval is \([-1, 1]\). 6. Integration leaves the radius equal to \(1\), but the added denominator factors can improve endpoint convergence.

a) \( F_1(x)=\sum_{m=1}^{\infty}\frac{x^m}{m}\), and \( F_2(x)=\sum_{j=2}^{\infty}\frac{x^j}{(j-1)j}\). b) \(F_0\): \((-1, 1)\); \(F_1\): \([-1, 1)\); \(F_2\): \([-1, 1]\). c) Repeated integration preserves the radius but can improve convergence at the endpoints.

- Test the two endpoints separately from the interior geometric-series calculation. - Use the geometric formula only where \(|x|<1\). - At \(x=1\), examine the sequence of ordinary partial sums. - Distinguish a boundary limit of the sum function from convergence of the endpoint numerical series.

1. The series is geometric with ratio \(-x\), so it converges when \(|x|<1\). At \(x=1\), its partial sums alternate between \(1\) and \(0\). At \(x=-1\), every term is \(1\). Thus both endpoints diverge. 2. The radius is \(1\), and the interval of convergence is \((-1, 1)\). 3. Inside that interval, \( F(x)=\frac{1}{1+x}\). Therefore, \( \lim_{x\to1^-}F(x)=\frac12\). 4. The closed-form identity was obtained only for \(|x|<1\). At \(x=1\), ordinary convergence is determined by the sequence of partial sums, which has no limit. Thus \(1/2\) is a boundary limit of the sum function, not the ordinary sum of the endpoint series.

a) Radius \(1\); interval \((-1, 1)\). b) \( F(x)=\frac{1}{1+x}\) for \(|x|<1\). c) \( \frac12\). d) The endpoint partial sums alternate between \(1\) and \(0\), so the numerical series diverges; \(1/2\) is only the one-sided limit of the interior sum function.

- Rewrite a term so that the known \(4^n\) bound becomes useful. - Compare the omitted terms with a geometric tail. - Check what happens to the comparison ratio at the endpoints.

1. For \(n\geq12\), \(|a_nx^n|=|a_n4^n|\left|\frac{x}{4}\right|^n\leq0.02\left|\frac{x}{4}\right|^n\). 2. The geometric comparison converges for \(|x|<4\), establishing absolute convergence. 3. At \(x=3\), \(|F(3)-P_{20}(3)|\leq0.02\sum_{n=21}^{\infty}\left(\frac{3}{4}\right)^n=0.08\left(\frac{3}{4}\right)^{21}\approx0.0001903\). 4. Taking absolute values at \(-3\) gives the same estimate. 5. At \(|x|=4\), the comparison ratio equals 1; the hypothesis neither proves that the terms approach zero nor controls whether the endpoint series converges.

a) Absolute convergence holds for \(|x|<4\). b) \(|F(3)-P_{20}(3)|\leq0.08\left(\frac{3}{4}\right)^{21}\approx0.0001903\). c) The same bound holds at \(x=-3\). d) No convergence conclusion at either endpoint follows.

- Measure each given point's distance from the center. - Convergence propagates inward from a point, while divergence propagates outward. - A single convergent point and a single divergent point usually provide bounds, not an exact radius.

1. The point \(x=-1\) is \(3\) units from the center. Since the series converges there, it must converge at every point closer to the center. Thus \(R\geq3\). 2. The point \(x=6\) is \(4\) units from the center. Since the series diverges there, the radius cannot exceed \(4\). Thus \(R\leq4\). 3. Therefore, \(3\leq R\leq4\). The data do not reveal whether either given point lies at an endpoint, so neither the exact radius nor all endpoint behavior is determined.

a) \(R\geq3\). b) \(R\leq4\). c) No. The only conclusion is \(3\leq R\leq4\).

- Where could conditional convergence occur relative to a power series' radius? - Compare absolute convergence at symmetric endpoints by taking absolute values of their terms. - Translate the three given behaviors into distances from the center. - Reflect the known boundary across the center to find the opposite boundary.

1. Conditional convergence at \(x=4\) puts that point on the boundary, so \(R=|4-c|\). 2. Absolute convergence at \(x=1\) means \(|1-c|<|4-c|\): the inequality must be strict because absolute convergence at one boundary would imply absolute convergence at the opposite boundary, contradicting conditional convergence at \(4\). Thus \(c<\frac{5}{2}\). 3. Divergence at \(x=-2\) requires \(|-2-c|\geq|4-c|\), giving \(c\geq1\). Together \(1\leq c<\frac{5}{2}\), and hence \(R=4-c\). 4. At \(c=1\), \(-2\) is the opposite boundary point and is known to diverge. For \(1<c<\frac{5}{2}\), \(-2\) lies strictly outside the radius. 5. The opposite boundary is \(2c-4\). It cannot converge absolutely, since absolute convergence at either boundary gives absolute convergence at both. At \(c=1\) it diverges as given; for \(1<c<\frac{5}{2}\) it may diverge or converge conditionally. 6. Every center in the stated range is possible. For each \(1\leq c<\frac{5}{2}\), take \(\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\left(\frac{x-c}{4-c}\right)^n\). It converges conditionally at \(4\), absolutely at \(1\), and diverges at \(-2\).

a) \(1\leq c<\frac{5}{2}\), with \(R=4-c\). b) \(-2\) is a boundary point when \(c=1\); it lies strictly outside for \(1<c<\frac{5}{2}\). c) The opposite boundary is \(2c-4\). It diverges at \(c=1\); otherwise it may diverge or converge conditionally, but not absolutely.

- Treat everything inside the nth power as the geometric ratio. - Compare the shape of a centered power-series convergence interval with the given geometric series. - Expand the rational function, not the original expression, about zero.

1. The geometric ratio is \(r=\frac{x^2-4}{3}\). Convergence requires \(|r|<1\), equivalent to \(1<x^2<7\). 2. Thus the original series converges on \((-\sqrt{7},-1)\cup(1,\sqrt{7})\) and sums to \(\frac{3}{7-x^2}\). 3. Its convergence set has two separate interior intervals. A power series centered at 0 instead converges throughout the interior of one interval centered there. 4. Rewriting the rational function as \(\frac{3}{7}\frac{1}{1-x^2/7}\) gives the Maclaurin series \(\frac{3}{7}\sum_{n=0}^{\infty}\left(\frac{x^2}{7}\right)^n\), which converges for \(|x|<\sqrt{7}\) and diverges at both endpoints. 5. The two series represent the same rational function only where both converge: \((-\sqrt{7},-1)\cup(1,\sqrt{7})\). In particular, the new series converges near 0 but the original one does not.

a) \((-\sqrt{7},-1)\cup(1,\sqrt{7})\). b) \(\frac{3}{7-x^2}\). c) The original series has a disconnected interior convergence set. d) \(\frac{3}{7}\sum_{n=0}^{\infty}\left(\frac{x^2}{7}\right)^n\), with interval \((-\sqrt{7},\sqrt{7})\); the series agree where \(1<|x|<\sqrt{7}\).

- What does a finite radius say about points farther from the center? - How far can positive integers be from one fixed real center? - Try to construct infinitely many inputs strictly inside the geometric series' interval of convergence.

1. If a power series had finite radius \(R\), it would diverge whenever \(|x-c|>R\). The given points eventually exceed any finite \(R\), a contradiction. 2. The distances of positive integers from any fixed center \(c\) are unbounded, so part a) applies. 3. Infinitely many points can lie within a fixed bounded interval and therefore within a finite radius. 4. The geometric series has radius 1 and converges at every \(x_k=1-\frac{1}{k}\) for integers \(k\geq2\). Those distinct points approach the boundary 1.

a) \(R=\infty\). b) The positive integers are arbitrarily far from any fixed center. c) An infinite set can stay within a bounded interval. d) The series \(\sum_{n=0}^{\infty}x^n\) has radius 1 and converges at \(x_k=1-\frac{1}{k}\), \(k\geq2\).