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Q 31.

Expert-verifiedFound in: Page 670

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 $\left(\frac{1}{2},\frac{5}{2}\right)$.

The given power series is:

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

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

$\underset{k\to \infty}{\mathrm{lim}}\left|\frac{{b}_{k+1}}{{b}_{k}}\right|=\underset{k\to \infty}{\mathrm{lim}}\left|\frac{{\left(-1\right)}^{k+1}\frac{\sqrt{k+1}}{{\left(k+1\right)}^{3}}{\left(x-\frac{3}{2}\right)}^{k+1}}{{\left(-1\right)}^{k}\frac{\sqrt{k}}{{k}^{3}}{\left(x-\frac{3}{2}\right)}^{k}}\right|\phantom{\rule{0ex}{0ex}}=\underset{k\to \infty}{\mathrm{lim}}\left|\left(-1\right)\frac{{k}^{3}}{{\left(k+1\right)}^{3}}\frac{\sqrt{k+1}}{\sqrt{k}}\left(x-\frac{3}{2}\right)\right|\phantom{\rule{0ex}{0ex}}=\underset{k\to \infty}{\mathrm{lim}}\left|x-\frac{3}{2}\right|{\left(\frac{k}{k+1}\right)}^{3}\sqrt{\frac{k+1}{k}}$

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 $\frac{1}{2}<x<\frac{5}{2}$.

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 $\sum _{k=1}^{\infty}{\left(-1\right)}^{k}\frac{\sqrt{k}}{{k}^{3}}{\left(x-\frac{3}{2}\right)}^{k}\mathrm{is}\left(\frac{1}{2},\frac{5}{2}\right)$.

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