Radius & Interval of Convergence - Calculus 2

1. For a fixed integer $k \ge 2$, find the radius of convergence of the series $$\sum_{n=0}^\infty \frac{(n!)^k}{(kn)!} x^n$$
   Answer: $R = k^k$ Apply the Ratio Test. $L = |x| \lim_{n \to \infty} \frac{((n+1)!)^k}{(k(n+1))!} \cdot \frac{(kn)!}{(n!)^k} = |x| \lim_{n \to \infty} \frac{(n+1)^k}{(kn+k)(kn+k-1)...(kn+1)}$. The numerator is a polynomial in $n$ with leading term $n^k$. The denominator is a product of $k$ linear terms, so it is a polynomial of degree $k$ with leading term $(kn)^k = k^k n^k$. The limit of the fraction is $1/k^k$. Thus, $L = \frac{|x|}{k^k}$. For convergence, $L<1$, which implies $|x|<k^k$. The radius of convergence is $R=k^k$.
2. The coefficients of a power series $\sum_{n=0}^{\infty} a_n x^n$ satisfy the recurrence relation $a_{n+1} = \frac{n}{2n+5} a_n$ for $n \ge 0$, with $a_0=1$. Find the radius of convergence of the series.
   Answer: $R = 2$ We can use the Ratio Test directly on the definition of the series. The limit of the ratio of consecutive terms is $L = \lim_{n \to \infty} \left| \frac{a_{n+1}x^{n+1}}{a_n x^n} \right| = |x| \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|$. Using the given recurrence relation, $\frac{a_{n+1}}{a_n} = \frac{n}{2n+5}$. The limit is $\lim_{n \to \infty} \frac{n}{2n+5} = \frac{1}{2}$. So, $L = \frac{1}{2}|x|$. The series converges when $L < 1$, which means $\frac{1}{2}|x| < 1$, or $|x| < 2$. The radius of convergence is $R=2$.
3. Consider the power series $\sum_{n=1}^{\infty} \frac{(-1)^n (x-2)^n}{n \cdot 3^n}$. What is the interval of convergence?
   Answer: $(-1, 5]$ Using the ratio test: $\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty} \frac{n}{(n+1) \cdot 3} = \frac{1}{3}$. So radius of convergence $R = 3$. The series converges for $|x-2| < 3$, or $-1 < x < 5$. At $x = -1$: $\sum_{n=1}^{\infty} \frac{(-1)^n(-3)^n}{n \cdot 3^n} = \sum_{n=1}^{\infty} \frac{1}{n}$, which diverges (harmonic series). At $x = 5$: $\sum_{n=1}^{\infty} \frac{(-1)^n \cdot 3^n}{n \cdot 3^n} = \sum_{n=1}^{\infty} \frac{(-1)^n}{n}$, which converges (alternating harmonic series). Therefore, the interval is $(-1, 5]$.
4. The power series $\sum_{n=0}^{\infty} c_n x^n$ has a radius of convergence of $R=5$. What is the radius of convergence of the series $\sum_{n=0}^{\infty} c_n (x^2 - 4)^n$?
   Answer: $R = 3$ The original series converges when its input is less than 5 in absolute value. In the second series, the input is $u = x^2 - 4$. Thus, the second series converges when $|u| < 5$, which translates to $|x^2 - 4| < 5$. This inequality is equivalent to $-5 < x^2 - 4 < 5$, or $-1 < x^2 < 9$. Since $x^2 \ge 0$, the left side $-1 < x^2$ is always true. The right side, $x^2 < 9$, implies $-3 < x < 3$. This interval is centered at $c=0$ and has a radius of $R=3$.
5. A power series $\sum_{n=0}^{\infty} a_n (x-3)^n$ is known to converge at $x=0$ and diverge at $x=8$. Which of the following statements must be true?
   Answer: The series converges at $x=5$. The series is centered at $c=3$. Since it converges at $x=0$, the distance from the center is $|0-3|=3$. This implies the radius of convergence $R$ is at least 3, so $R \ge 3$. Since it diverges at $x=8$, the distance from the center is $|8-3|=5$. This implies the radius of convergence $R$ is at most 5, so $R \le 5$. Therefore, we know $3 \le R \le 5$. The series must converge for all $x$ such that $|x-3| < R$. For $x=5$, the distance is $|5-3|=2$. Since $2 < 3 \le R$, the series must converge at $x=5$. The other options are not guaranteed: at $x=7$ (distance 4), convergence depends on whether $R>4$; at $x=-2$ (distance 5) and $x=-1$ (distance 4), convergence is not certain.
6. For the power series $\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}$, which statement correctly describes its interval of convergence?
   Answer: The series converges for all real numbers $x$ Using the ratio test on the series $\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}$: $\lim_{n \to \infty} \left|\frac{x^{2(n+1)}(2n)!}{x^{2n}(2(n+1))!}\right| = \lim_{n \to \infty} \frac{x^2}{(2n+2)(2n+1)} = 0$ for any finite $x$. Since this limit is 0 (less than 1) for all $x$, the radius of convergence is $R = \infty$. This series is actually $\cos(x)$, which converges for all real $x$. Choice A assumes finite radius, choice C gives the wrong interval, and choice D uses confusing notation that essentially says the same as choice B but in a misleading way.
7. Determine the interval of convergence for the power series $$\sum_{n=1}^{\infty} \frac{(-1)^n (x-2)^n}{n \cdot 4^n}$$
   Answer: $(-2, 6]$ Using the Ratio Test, we find the limit $\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{(x-2)^{n+1}}{(n+1)4^{n+1}} \cdot \frac{n 4^n}{(x-2)^n} \right| = \left| \frac{x-2}{4} \right|$. The series converges when $\left| \frac{x-2}{4} \right| < 1$, which simplifies to $|x-2| < 4$. The radius of convergence is $R=4$, and the open interval is $-2 < x < 6$. We must test the endpoints. At $x=6$, the series is $\sum \frac{(-1)^n}{n}$, the alternating harmonic series, which converges. At $x=-2$, the series is $\sum \frac{(-1)^n (-4)^n}{n \cdot 4^n} = \sum \frac{1}{n}$, the harmonic series, which diverges. Therefore, the interval of convergence is $(-2, 6]$.
8. Let $f(x) = \sum_{n=0}^{\infty} \frac{x^n}{2^n}$ and $g(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^n}{n!}$. What is the radius of convergence of the power series for the sum $h(x) = f(x) + g(x)$?
   Answer: $R = 2$ The series for $f(x)$ is a geometric series with ratio $x/2$. It converges for $|x/2| < 1$, which means $|x|<2$. Its radius of convergence is $R_f = 2$. The series for $g(x)$ is the well-known series for $e^{-x}$, which has a radius of convergence $R_g = \infty$. The radius of convergence of the sum of two power series is the minimum of their individual radii of convergence. Therefore, the radius for $h(x)$ is $R_h = \min(R_f, R_g) = \min(2, \infty) = 2$.
9. The power series $f(x) = \sum_{n=1}^{\infty} \frac{(x-4)^n}{n^2}$ has an interval of convergence of $[3, 5]$. What is the interval of convergence for the series representing its derivative, $f'(x)$?
   Answer: $[3, 5)$ Differentiation of a power series does not change its radius of convergence. The series for $f(x)$ is centered at $c=4$ and has $R=1$, so the series for $f'(x)$ also has $R=1$ and converges on $(3, 5)$. We must re-check the endpoints for the derivative series, $f'(x) = \sum_{n=1}^{\infty} \frac{n(x-4)^{n-1}}{n^2} = \sum_{n=1}^{\infty} \frac{(x-4)^{n-1}}{n}$. At $x=5$, the series is $\sum \frac{1}{n}$, the harmonic series, which diverges. At $x=3$, the series is $\sum \frac{(-1)^{n-1}}{n}$, the alternating harmonic series, which converges. Thus, the interval of convergence for $f'(x)$ is $[3, 5)$.
10. Find the interval of convergence for the power series $$\sum_{n=2}^{\infty} \frac{x^n}{\ln n}$$
    Answer: $[-1, 1)$ Using the Ratio Test, $\lim_{n \to \infty} \left| \frac{x^{n+1}}{\ln(n+1)} \cdot \frac{\ln n}{x^n} \right| = |x| \lim_{n \to \infty} \frac{\ln n}{\ln(n+1)}$. By L'Hôpital's Rule, this limit is 1. So the series converges for $|x| < 1$, meaning $R=1$. We test the endpoints. At $x=1$, the series is $\sum \frac{1}{\ln n}$. Since $\ln n < n$ for $n \ge 1$, $\frac{1}{\ln n} > \frac{1}{n}$. By direct comparison with the divergent harmonic series, $\sum \frac{1}{\ln n}$ diverges. At $x=-1$, the series is $\sum \frac{(-1)^n}{\ln n}$. This is a convergent alternating series since $1/\ln n$ is positive, decreasing, and approaches 0. Thus, the interval of convergence is $[-1, 1)$.
11. The power series $\sum_{n=2}^{\infty} \frac{x^n}{n(\ln n)^2}$ has interval of convergence $I$. Which statement about $I$ is correct?
    Answer: $I = [-1, 1]$ and the series converges absolutely at both endpoints Using ratio test: $\lim_{n \to \infty} \left|\frac{n(\ln n)^2}{(n+1)(\ln(n+1))^2}\right| = 1$, so $R = 1$. At $x = 1$: $\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^2}$ converges by integral test since $\int_2^{\infty} \frac{1}{x(\ln x)^2} dx = \left[-\frac{1}{\ln x}\right]_2^{\infty} = \frac{1}{\ln 2}$. At $x = -1$: $\sum_{n=2}^{\infty} \frac{(-1)^n}{n(\ln n)^2}$ converges absolutely since $\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^2}$ converges. Therefore $I = [-1, 1]$ with absolute convergence at both endpoints.
12. If $\sum_{n=0}^{\infty} a_n x^n$ has radius of convergence $R = 3$, what is the radius of convergence of $\sum_{n=0}^{\infty} a_n \frac{(x-1)^{2n}}{2^n}$?
    Answer: $\sqrt{6}$ When you encounter power series with modified forms, you need to identify how the transformation affects the radius of convergence. The key is recognizing that the new series can be rewritten in terms of a substitution into the original series. Starting with $\sum_{n=0}^{\infty} a_n x^n$ having radius of convergence $R = 3$, let's analyze $\sum_{n=0}^{\infty} a_n \frac{(x-1)^{2n}}{2^n}$. First, rewrite this as $\sum_{n=0}^{\infty} a_n \left(\frac{(x-1)^2}{2}\right)^n$. This is exactly the original series with $u = \frac{(x-1)^2}{2}$ substituted for $x$. Since the original series converges when $|x| < 3$, the transformed series converges when $|u| < 3$, which means $\left|\frac{(x-1)^2}{2}\right| < 3$. Solving this inequality: $\frac{(x-1)^2}{2} < 3$, so $(x-1)^2 < 6$, giving us $|x-1| < \sqrt{6}$. Therefore, the radius of convergence is $\sqrt{6}$, which is answer D. Looking at the wrong answers: A) $6$ likely comes from forgetting to take the square root when solving $(x-1)^2 < 6$. B) $\sqrt{3}$ results from incorrectly handling the factor of 2 in the denominator. C) $3$ suggests ignoring the transformation entirely and assuming the radius stays the same. Strategy tip: For transformed power series, always substitute the new variable into the convergence condition of the original series. The algebra of solving the resulting inequality determines the new radius—don't try to memorize transformation rules.
13. Find the interval of convergence of the series $$\sum_{n=1}^{\infty} \frac{(2x-5)^n}{n^2}$$
    Answer: $[2, 3]$ Let $u = 2x-5$. The series becomes $\sum_{n=1}^{\infty} \frac{u^n}{n^2}$. The radius of convergence for this series in $u$ is $R_u=1$, converging for $|u|<1$. Substituting back, we have $|2x-5|<1$, which is $-1 < 2x-5 < 1$. Adding 5 gives $4 < 2x < 6$, and dividing by 2 gives $2 < x < 3$. Now we check the endpoints. At $x=2$, $u=-1$ and the series is $\sum \frac{(-1)^n}{n^2}$, which converges absolutely. At $x=3$, $u=1$ and the series is $\sum \frac{1}{n^2}$, a convergent p-series. Both endpoints are included, so the interval is $[2, 3]$.
14. What is the radius of convergence $R$ for the series $$\sum_{n=1}^{\infty} \frac{(n!)^2}{(2n)!} (x-1)^n$$?
    Answer: $R = 4$ We use the Ratio Test. $L = \lim_{n \to \infty} \left| \frac{((n+1)!)^2 (x-1)^{n+1}}{(2(n+1))!} \cdot \frac{(2n)!}{(n!)^2 (x-1)^n} \right| = |x-1| \lim_{n \to \infty} \frac{(n+1)^2}{(2n+2)(2n+1)}$. The limit of the rational expression is $\lim_{n \to \infty} \frac{n^2+2n+1}{4n^2+6n+2} = \frac{1}{4}$. Thus, $L = \frac{1}{4}|x-1|$. For convergence, we need $L < 1$, which implies $\frac{1}{4}|x-1| < 1$, or $|x-1| < 4$. The radius of convergence is $R=4$.
15. Consider the power series $\sum_{n=0}^{\infty} a_n x^n$ where $a_n = \begin{cases} \frac{1}{2^n} & \text{if } n \text{ is even} \\ \frac{1}{3^n} & \text{if } n \text{ is odd} \end{cases}$. What is the radius of convergence?
    Answer: $R = 2$ When you encounter a power series with coefficients that follow different patterns for even and odd terms, you need to use the root test (Cauchy-Hadamard theorem) rather than the ratio test, since the coefficients don't follow a single pattern. For the radius of convergence, we use $R = \frac{1}{\limsup_{n \to \infty} \sqrt[n]{|a_n|}}$. Since the coefficients alternate between two different formulas, we need to find $\limsup_{n \to \infty} \sqrt[n]{|a_n|}$ by examining both subsequences. For even $n$: $\sqrt[n]{a_n} = \sqrt[n]{\frac{1}{2^n}} = \frac{1}{2}$ For odd $n$: $\sqrt[n]{a_n} = \sqrt[n]{\frac{1}{3^n}} = \frac{1}{3}$ The $\limsup$ is the larger of these two values, which is $\frac{1}{2}$. Therefore, $R = \frac{1}{1/2} = 2$. Choice A ($R = \frac{\sqrt{6}}{2}$) likely comes from incorrectly trying to find some average or geometric mean of the two patterns. Choice B ($R = 3$) results from mistakenly taking the reciprocal of $\frac{1}{3}$ and ignoring the even terms. Choice C ($R = \sqrt{6}$) might arise from incorrectly multiplying the individual radii $2 \times 3 = 6$ and taking the square root. Study tip: When coefficients follow different patterns for even and odd indices, always use the root test and remember that $\limsup$ takes the larger limiting value. The series converges only where both subsequences would converge individually.
16. Find the radius of convergence for the power series $$\sum_{n=1}^\infty \left(1 + \frac{1}{n}\right)^{n \ln n} x^n$$
    Answer: $R = 1$ Using the Root Test, we evaluate $L = \lim_{n \to \infty} \sqrt[n]{\left| \left(1 + \frac{1}{n}\right)^{n \ln n} x^n \right|} = |x| \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^{\ln n}$. To evaluate the limit, let $y = (1 + 1/n)^{\ln n}$, so $\ln y = (\ln n) \ln(1+1/n)$. Using the approximation $\ln(1+u) \approx u$ for small $u$, we get $\ln y \approx (\ln n)(1/n) = \frac{\ln n}{n}$. As $n \to \infty$, $\lim \frac{\ln n}{n} = 0$. So $\lim \ln y = 0$, which means $\lim y = e^0 = 1$. Thus, $L = |x| \cdot 1 = |x|$. For convergence, $|x|<1$, so the radius of convergence is $R=1$.
17. Let $R$ be the radius of convergence of the power series $\sum_{n=0}^{\infty} a_n x^n$. If it is known that the series of coefficients $\sum_{n=0}^{\infty} a_n$ converges, what is the strongest conclusion that can be made about $R$?
    Answer: $R \ge 1$ The series $\sum_{n=0}^{\infty} a_n$ is the power series $\sum_{n=0}^{\infty} a_n x^n$ evaluated at $x=1$. The fact that the power series converges for a specific value $x=1$ means that its radius of convergence $R$ must be at least the distance from the center (c=0) to that point. So, $R \ge |1-0| = 1$. It is possible that $R=1$ (e.g., for $a_n = (-1)^n/n$) or that $R>1$ (e.g., for $a_n = (1/2)^n$). The only certain conclusion is $R \ge 1$.
18. For the power series $\sum_{n=1}^{\infty} \frac{(-1)^n x^n}{n^{\alpha}}$ where $\alpha > 0$, the interval of convergence changes behavior at a critical value of $\alpha$. What is this critical value and how does the interval change?
    Answer: At $\alpha = 1$, the interval changes from $(-1,1]$ to $[-1,1]$ When analyzing power series convergence, you need to find the radius of convergence first, then test the endpoints separately. The behavior at endpoints depends critically on the series coefficients. For $\sum_{n=1}^{\infty} \frac{(-1)^n x^n}{n^{\alpha}}$, the radius of convergence is $R = 1$ (using the ratio test), so you must check convergence at $x = -1$ and $x = 1$. At $x = 1$: The series becomes $\sum_{n=1}^{\infty} \frac{(-1)^n}{n^{\alpha}}$, which is an alternating series. By the alternating series test, this converges for any $\alpha > 0$ since $\frac{1}{n^{\alpha}}$ decreases to zero. At $x = -1$: The series becomes $\sum_{n=1}^{\infty} \frac{(-1)^n (-1)^n}{n^{\alpha}} = \sum_{n=1}^{\infty} \frac{1}{n^{\alpha}}$, which is a p-series. This converges only when $\alpha > 1$. Therefore: For $0 < \alpha \leq 1$, the interval is $(-1,1]$. For $\alpha > 1$, the interval is $[-1,1]$. The critical transition occurs at $\alpha = 1$, making C correct. Choice A incorrectly suggests the interval starts as $(-1,1)$ and that $x = 1$ is excluded after the transition. Choice B uses the wrong critical value and wrong endpoint behavior. Choice D uses $\alpha = 2$ as the critical value and incorrectly suggests $x = 1$ was initially excluded. Study tip: For power series endpoint analysis, always check alternating series behavior at one endpoint and p-series behavior at the other. The critical value $\alpha = 1$ frequently appears because it's where p-series convergence begins.
19. The Maclaurin series for $\cos(x)$ is $\sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!}$, which has an infinite radius of convergence. What is the interval of convergence for the Taylor series of $f(x)=\cos(x-1)$ centered at $c=1$?
    Answer: $(-\infty, \infty)$ The Taylor series for $\cos(x-1)$ centered at $c=1$ is obtained by substituting $(x-1)$ for $x$ in the Maclaurin series for $\cos(x)$. This gives $\sum_{n=0}^\infty \frac{(-1)^n (x-1)^{2n}}{(2n)!}$. Since the original series for $\cos(u)$ converges for all real numbers $u$, the substituted series will converge for all values of $x-1$, which means it converges for all real numbers $x$. The substitution simply shifts the center of the series expansion; it does not change the radius of convergence. Therefore, the interval of convergence is $(-\infty, \infty)$.
20. The power series $\sum_{n=0}^{\infty} \frac{(3x-6)^n}{\sqrt{n+1}}$ has interval of convergence $I$. Which interval represents $I$?
    Answer: $\left[\frac{5}{3}, \frac{7}{3}\right)$ Rewrite as $\sum_{n=0}^{\infty} \frac{3^n(x-2)^n}{\sqrt{n+1}}$. Using ratio test: $\lim_{n \to \infty} \left|\frac{3\sqrt{n+1}}{\sqrt{n+2}}\right| = 3$, so radius is $R = \frac{1}{3}$ around $x = 2$. Series converges for $|x-2| < \frac{1}{3}$, i.e., $\frac{5}{3} < x < \frac{7}{3}$. At $x = \frac{5}{3}$: $\sum_{n=0}^{\infty} \frac{(-1)^n}{\sqrt{n+1}}$ converges by alternating series test. At $x = \frac{7}{3}$: $\sum_{n=0}^{\infty} \frac{1}{\sqrt{n+1}}$ diverges (p-series with $p = \frac{1}{2} < 1$). Therefore $I = \left[\frac{5}{3}, \frac{7}{3}\right)$.

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