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Expert-verified Found in: Page 670 ### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861 # Find the interval of convergence for each power series in Exercises 21–48. If the interval of convergence is finite, be sure to analyze the convergence at the endpoints.$\sum _{k=0}^{\infty }\frac{{\left(k!\right)}^{2}}{\left(2k!\right)}{\left(x+2\right)}^{k}$

The interval of convergence for power series is $\left(-4,0\right)$.

See the step by step solution

## Step 1. Given information.

The given power series is:

$\sum _{k=0}^{\infty }\frac{{\left(k!\right)}^{2}}{\left(2k!\right)}{\left(x+2\right)}^{k}$

## Step 2. Find the interval of convergence.

Let us assume ${b}_{k}=\frac{{\left(k!\right)}^{2}}{\left(2k!\right)}{\left(x+2\right)}^{k}$ therefore

${b}_{k+1}=\frac{{\left[\left(k+1\right)!\right]}^{2}}{\left[2\left(k+1\right)\right]!}{\left(x+2\right)}^{k+1}$

The ratio for the absolute convergence is

Here the limit role="math" $\frac{1}{2}\left|x+2\right|$ is So, by the ratio test of absolute convergence, we know that series will converge absolutely when $\frac{1}{2}\left|x+2\right|<1$ that is $-4.

## Step 3. Find the interval of convergence.

Now, since the intervals are finite so we analyze the behavior of the series at the endpoints.

So, when $x=-4$

$\sum _{k=0}^{\infty }\frac{{\left(k!\right)}^{2}}{\left(2k!\right)}{\left(x+2\right)}^{k}=\sum _{k=0}^{\infty }\frac{{\left(k!\right)}^{2}}{\left(2k!\right)}{\left(-4+2\right)}^{k}\phantom{\rule{0ex}{0ex}}=\sum _{k=0}^{\infty }\frac{{\left(k!\right)}^{2}}{\left(2k!\right)}{\left(-2\right)}^{k}$

The result is just a constant multiple of the arithmetic series, which diverges conditionally.

So, when $x=0$

$\sum _{k=0}^{\infty }\frac{{\left(k!\right)}^{2}}{\left(2k!\right)}{\left(x+2\right)}^{k}=\sum _{k=0}^{\infty }\frac{{\left(k!\right)}^{2}}{\left(2k!\right)}{\left(0+2\right)}^{k}\phantom{\rule{0ex}{0ex}}=\sum _{k=0}^{\infty }\frac{{\left(k!\right)}^{2}}{\left(2k!\right)}{\left(2\right)}^{k}$

The result is the alternating multiple of the harmonic series, which diverges.

Therefore, the interval of convergence of the power series $\sum _{k=0}^{\infty }\frac{{\left(k!\right)}^{2}}{\left(2k!\right)}{\left(x+2\right)}^{k}$ is $\left(-4,0\right)$. ### Want to see more solutions like these? 