Free math worksheets

- 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 \(R=3\).

\(R=3\).

- 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.

- Keep the exponent \(2n\) intact when forming the ratio. - After solving the ratio inequality, test both endpoints in the original series.

1. Let \( a_n=\frac{(x-2)^{2n}}{n\,9^n}\). Then \( \left|\frac{a_{n+1}}{a_n}\right|=\frac{|x-2|^2}{9}\cdot\frac{n}{n+1}\to\frac{|x-2|^2}{9}\). 2. The ratio test gives convergence when \( \frac{|x-2|^2}{9}<1\), or \(|x-2|<3\). Thus the radius is \(3\). 3. At \(x=-1\) and \(x=5\), the series becomes \(\sum_{n=1}^{\infty}\frac{1}{n}\), which diverges. Therefore, the interval is \((-1, 5)\).

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

- Form the ratio after replacing \(n\) by \(n+1\) in every factor. - Simplify the factorial quotient before taking the limit. - A ratio limit of \(0\) works for every fixed value of \(x\).

1. Let \(u_n=\frac{n^2(x+1)^n}{n!}\). The ratio of consecutive absolute values is \( \left|\frac{u_{n+1}}{u_n}\right|=\frac{(n+1)^2|x+1|}{(n+1)n^2}=\frac{n+1}{n^2}|x+1|\). 2. As \(n\to\infty\), \(\frac{n+1}{n^2}\to0\), so the ratio limit is \(0\) for every real \(x\). 3. The series therefore converges absolutely for every real number. Its radius is infinite, and its interval of convergence is \((-\infty, \infty)\).

Radius: \(R=\infty\). Interval of convergence: \((-\infty, \infty)\).

- 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 polynomial factor into each term of the series. - Identify which part of the general term changes with \(n\) exponentially. - Recheck the endpoints even when the radius is unchanged.

1. Multiplication gives \( G(x)=\sum_{n=0}^{\infty}\frac{(x-2)^{n+4}}{3^n}\). 2. The underlying geometric ratio is still \(\frac{x-2}{3}\), so convergence requires \(|x-2|<3\). Thus \(R=3\) and the preliminary interval is \((-1, 5)\). 3. At \(x=-1\) and \(x=5\), the terms of \(G\) do not approach \(0\), so both endpoints diverge. 4. Multiplying by a fixed polynomial changes only finitely many initial powers and adds a finite factor for each fixed \(x\); it does not alter the geometric ratio controlling convergence away from the endpoints.

a) \( G(x)=\sum_{n=0}^{\infty}\frac{(x-2)^{n+4}}{3^n}\). b) Radius \(R=3\); interval \((-1, 5)\). c) The same geometric ratio \(\frac{x-2}{3}\) controls convergence.

- 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 \(x\in(-3, 7)\).

- 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 the convergence set of each component separately. - The sum converges only where both component series converge. - At each boundary of the smaller interval, substitute the endpoint into the component that controls the radius.

1. Rewrite the first component as \( \sum_{n=1}^{\infty}\frac{1}{n}\left(\frac{(x+1)^2}{9}\right)^n\). It converges when \(\frac{(x+1)^2}{9}<1\), so \(|x+1|<3\). 2. Rewrite the second component as \( \sum_{n=1}^{\infty}\frac{1}{n^2}\left(\frac{x+1}{4}\right)^n\). It converges for \(|x+1|\leq4\), including both endpoints because the endpoint series are absolutely convergent \(p\)-series. 3. Both components must converge, so the first component gives the smaller radius. The center is \(-1\) and \(R=3\), giving the open interval \((-4, 2)\). 4. At \(x=-4\), the first component becomes \( \sum_{n=1}^{\infty}\frac{1}{n}\), so the total series diverges. 5. At \(x=2\), the first component again becomes \( \sum_{n=1}^{\infty}\frac{1}{n}\), so the total series diverges. 6. Therefore, the interval of convergence is \((-4, 2)\).

Radius: \(R=3\). Interval: \((-4, 2)\). The first component determines the radius, and the 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.

- Insert the factor \(4^n\) from the hypothesis into each term. - Bound the omitted tail by a geometric series. - Notice that the comparison ratio becomes \(1\) at the endpoints, where the argument stops working. - In part d), explain why a comparison with ratio \(1\) gives no endpoint conclusion.

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. When \(|x|<4\), the comparison series is geometric with ratio less than \(1\). Therefore, \(F(x)\) converges absolutely. 3. At \(x=3\), \( |F(3)-P_{20}(3)|\leq0.02\sum_{n=21}^{\infty}\left(\frac34\right)^n\). Thus \( |F(3)-P_{20}(3)|\leq0.08\left(\frac34\right)^{21}\approx0.0001903\). 4. At \(x=-3\), absolute values again produce the ratio \(\frac34\), so the same estimate applies. 5. At \(|x|=4\), the hypothesis only gives \(|a_nx^n|\leq0.02\). It does not force the terms to approach \(0\) and does not provide a convergent comparison series. Therefore, either endpoint may converge or diverge.

a) The series converges absolutely for every \(|x|<4\). b) \( |F(3)-P_{20}(3)|\leq0.08\left(\frac34\right)^{21}\approx0.0001903\). c) The same bound holds at \(x=-3\). d) No endpoint conclusion follows from the estimate alone.

- 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 \(3\leq R\leq4\).

- Conditional convergence places a point on the boundary of a power series. - Translate the other two facts into distance inequalities from the unknown center. - Find the opposite boundary by reflecting \(4\) across the center.

1. Conditional convergence at \(x=4\) means that \(4\) is a boundary point. Therefore, \( R=|4-c|\). 2. Absolute convergence at \(x=1\) requires \( |1-c|<|4-c|\). Squaring both sides and simplifying gives \(c<\frac52\). 3. Divergence at \(x=-2\) means that this point is not strictly inside the radius: \( |-2-c|\geq|4-c|\). Squaring and simplifying gives \(c\geq1\). 4. Hence \( 1\leq c<\frac52\). In this range, \(c<4\), so \(R=4-c\). 5. When \(c=1\), the radius is \(3\), so \(x=-2\) is the boundary opposite \(x=4\). When \(c>1\), \(x=-2\) lies strictly outside the radius. 6. The opposite boundary is \( c-R=c-(4-c)=2c-4\). It is known to diverge only in the case \(c=1\), when it equals \(-2\). For other centers, the given information does not determine its endpoint behavior.

a) \( 1\leq c<\frac52\), with \(R=4-c\). b) \(x=-2\) is a boundary point when \(c=1\); it is strictly outside when \(1<c<\frac52\). c) The opposite boundary is \(2c-4\). Its behavior is undetermined except when \(c=1\), where it diverges.

- Treat the entire expression inside the nth power as the geometric ratio. - Compare the shape of the convergence set with the shape required for a centered power series. - Rewrite the closed-form sum using a different geometric ratio for the Maclaurin expansion. - For part d), expand the closed-form sum about \(0\) and test the new endpoints separately.

1. The geometric ratio is \( r=\frac{x^2-4}{3}\). Convergence requires \(|r|<1\), so \( |x^2-4|<3\). 2. Solving \( -3<x^2-4<3\) gives \(1<x^2<7\). Therefore, the displayed series converges on \( (-\sqrt7, -1)\cup(1, \sqrt7)\). At all four boundary points, the ratio is \(1\) or \(-1\), so the terms do not approach \(0\). 3. Its geometric sum is \( G(x)=\frac{1}{1-\frac{x^2-4}{3}}=\frac{3}{7-x^2}\). 4. An ordinary power series centered at \(0\) has one interval of convergence symmetric about \(0\). The disconnected convergence set above shows that the displayed expression is geometric in \(x^2-4\), not a power series in powers of \(x\). 5. Rewriting the sum gives \( \frac{3}{7-x^2}=\frac37\frac{1}{1-\frac{x^2}{7}}=\frac37\sum_{n=0}^{\infty}\left(\frac{x^2}{7}\right)^n\). This Maclaurin series converges for \(|x|<\sqrt7\), and its endpoints are excluded because the terms do not approach \(0\).

a) \( (-\sqrt7, -1)\cup(1, \sqrt7)\). b) \( G(x)=\frac{3}{7-x^2}\). c) Its convergence set is disconnected, unlike the interval required for an ordinary centered power series. d) \( \frac{3}{7-x^2}=\frac37\sum_{n=0}^{\infty}\left(\frac{x^2}{7}\right)^n\), with interval \((-\sqrt7, \sqrt7)\).

- Recall what a finite radius says about every point farther than that radius from the center. - Compare the distances of the positive integers from a fixed real number \(c\). - For the counterexample, choose infinitely many inputs strictly inside the interval of a geometric series. - For part d), choose a sequence of points inside \((-1, 1)\) that approaches \(1\).

1. A power series with finite radius \(R\) diverges at every point whose distance from its center is greater than \(R\). 2. Suppose the radius were finite. Because \(|x_k-c|\to\infty\), some \(x_k\) would satisfy \(|x_k-c|>R\). The series would then have to diverge at that point, contradicting the hypothesis. Therefore, no finite radius is possible, and \(R=\infty\). 3. For any fixed real center \(c\), the distances \(|m-c|\) become arbitrarily large as the positive integer \(m\) increases. Thus convergence at every positive integer satisfies the condition from part a). 4. Infinitely many points can all remain inside one bounded convergence interval. Their number does not force their distances from the center to be unbounded. 5. For example, \(\sum_{n=0}^{\infty}x^n\) has radius \(1\), but it converges at every point \(x_k=1-\frac1k\) for \(k\geq2\). These distinct points approach the finite boundary point \(1\).

a) The radius is infinite because convergence occurs at distances from the center that exceed every finite number. b) Positive integers have unbounded distance from any fixed center, so convergence at all of them forces infinite radius. c) Infinitely many convergence points may remain in a bounded interval. d) \(\sum_{n=0}^{\infty}x^n\) has radius \(1\) and converges at \(x_k=1-\frac1k\) for every \(k\geq2\).