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Q 31.

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.$\sum _{k=1}^{\infty }{\left(-1\right)}^{k}\frac{\sqrt{k}}{{k}^{3}}{{\left(x-\frac{3}{2}\right)}^{k}}^{}$

The interval of convergence for power series is $\left(\frac{1}{2},\frac{5}{2}\right)$.

See the step by step solution

## Step 1. Given information.

The given power series is:

$\sum _{k=1}^{\infty }{\left(-1\right)}^{k}\frac{\sqrt{k}}{{k}^{3}}{\left(x-\frac{3}{2}\right)}^{k}$

## Step 2. Find the interval of convergence.

Let us assume ${b}_{k}={\left(-1\right)}^{k}\frac{\sqrt{k}}{{k}^{3}}{\left(x-\frac{3}{2}\right)}^{k}$ therefore

${b}_{k+1}={\left(-1\right)}^{k+1}\frac{\sqrt{k+1}}{{\left(k+1\right)}^{3}}{\left(x-\frac{3}{2}\right)}^{k+1}$

The ratio for the absolute convergence is

Here the limit is $\left|x-\frac{3}{2}\right|$ So, by the ratio test of absolute convergence, we know that series will converge absolutely when $\left|x-\frac{3}{2}\right|<1$ that is $\frac{1}{2}.

## 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 $x=\frac{1}{2}$

$\sum _{k=1}^{\infty }{\left(-1\right)}^{k}\frac{\sqrt{k}}{{k}^{3}}{\left(x-\frac{3}{2}\right)}^{k}=\sum _{k=1}^{\infty }{\left(-1\right)}^{k}\frac{\sqrt{k}}{{k}^{3}}{\left(\frac{1}{2}-\frac{3}{2}\right)}^{k}\phantom{\rule{0ex}{0ex}}=\sum _{k=1}^{\infty }{\left(-1\right)}^{k}\frac{\sqrt{k}}{{k}^{3}}{\left(-1\right)}^{k}\phantom{\rule{0ex}{0ex}}=\sum _{k=1}^{\infty }\frac{\sqrt{k}}{{k}^{3}}{\left(-1\right)}^{2k}$

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

So, when $x=\frac{5}{2}$

$\sum _{k=1}^{\infty }{\left(-1\right)}^{k}\frac{\sqrt{k}}{{k}^{3}}{\left(x-\frac{3}{2}\right)}^{k}=\sum _{k=1}^{\infty }{\left(-1\right)}^{k}\frac{\sqrt{k}}{{k}^{3}}{\left(\frac{5}{2}-\frac{3}{2}\right)}^{k}\phantom{\rule{0ex}{0ex}}=\sum _{k=1}^{\infty }{\left(-1\right)}^{k}\frac{\sqrt{k}}{{k}^{3}}{\left(1\right)}^{k}\phantom{\rule{0ex}{0ex}}=\sum _{k=1}^{\infty }\frac{\sqrt{k}}{{k}^{3}}{\left(-1\right)}^{k}$The result is just a constant multiple of the harmonic series which converges conditionally.

Therefore, the interval of convergence of the power series is $\sum _{k=1}^{\infty }{\left(-1\right)}^{k}\frac{\sqrt{k}}{{k}^{3}}{\left(x-\frac{3}{2}\right)}^{k}\mathrm{is}\left(\frac{1}{2},\frac{5}{2}\right)$.