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### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

# In problems 1-6, determine the convergence set of the given power series. $\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\frac{{\mathbf{n}}^{\mathbf{2}}}{{\mathbf{2}}^{\mathbf{n}}}{\mathbf{\left(}\mathbf{x}\mathbf{+}\mathbf{2}\mathbf{\right)}}^{{\mathbf{n}}}$

The set is $x\in \left(-4,0\right)$.

See the step by step solution

## Step 1:To Find the Radius of convergence

Use The Ratio Test to find the radius of convergence. In this case,

${a}_{n}=\frac{{n}^{2}}{{2}^{n}}$

Therefore,

${a}_{n+1}=\frac{{\left(n+1\right)}^{2}}{{2}^{n+1}}$

and

$\begin{array}{c}\underset{n\to \infty }{\mathrm{lim}}|\frac{{a}_{n}}{{a}_{n+1}}|=\underset{n\to \infty }{\mathrm{lim}}\mid \frac{\frac{{n}^{2}}{{2}^{n}}}{\frac{{\left(n+1\right)}^{2}}{{2}^{n+1}}\mid }\\ =\underset{n\to \infty }{\mathrm{lim}}|\frac{{n}^{2}{2}^{n+1}}{{2}^{n}{\left(n+1\right)}^{2}}|\\ =\underset{n\to \infty }{\mathrm{lim}}|\frac{{n}^{2}}{{\left(n+1\right)}^{2}}×\frac{{2}^{n+1}}{{2}^{n}}|\\ =\underset{n\to \infty }{\mathrm{lim}}|\frac{{n}^{2}}{{\left(n+1\right)}^{2}}×{2}^{n+1-n}|\\ =\underset{n\to \infty }{\mathrm{lim}}|\frac{{n}^{2}}{{\left(n+1\right)}^{2}}×2|\end{array}$

Simplify further.

$\begin{array}{c}\underset{n\to \infty }{\mathrm{lim}}|\frac{{a}_{n}}{{a}_{n+1}}|=\underset{n\to \infty }{\mathrm{lim}}\frac{2{n}^{2}}{{n}^{2}+2n+1}\\ \stackrel{}{=}2\underset{n\to \infty }{\mathrm{lim}}\frac{{n}^{2}}{{n}^{2}\left(1+\frac{2}{n}+\frac{1}{{n}^{2}}\right)}\\ \stackrel{}{=}2\underset{n\to \infty }{\mathrm{lim}}\frac{1}{1+\frac{2}{n}+\frac{1}{{n}^{2}}}\\ =2\text{×}\frac{1}{1}\\ =2\end{array}$

Therefore, the radius of convergence is$r=2$ , which means that the series converges for $|x+2|<2$or, $-2that is$x\in \left(-4,0\right)$.

(1): Remove the absolute value, since $\frac{{n}^{2}}{{\left(n+1\right)}^{2}}>0$always.

(2): Strip out the largest power of in the denominator and then cancel it

(3): Here $\frac{2}{n}\to 0,n\to \infty$and $\frac{1}{{n}^{2}}\to 0,n\to \infty$.

## Step 2: Check for convergence

Now, we need to check whether the boundary points $x=-4$and $x=0$are also in the set. We do that, by substituting$x=0$ and $x=-4$in the series.$x=-4$

Substituting $-4$for gives:

$\begin{array}{c}\sum _{n=0}^{\infty }\frac{{n}^{2}}{{2}^{n}}{\left(-4+2\right)}^{n}=\sum _{n=0}^{\infty }\frac{{n}^{2}}{{2}^{n}}{\left(-2\right)}^{n}\\ =\sum _{n=0}^{\infty }\frac{{n}^{2}}{{2}^{n}}×{2}^{n}×{\left(-1\right)}^{n}\\ =\sum _{n=0}^{\infty }{\left(-1\right)}^{n}{n}^{2}\end{array}$

This is an alternating series. Therefore, use the alternating series test, to check whether it converges. First, check if

$\underset{n\to \infty }{\mathrm{lim}}{a}_{n}=0$

In this case,${a}_{n}={n}^{2}$and it follows that.$\underset{n\to \infty }{\mathrm{lim}}{n}^{2}=\infty \ne 0$

(Also,${a}_{n}={n}^{2}$ is not monotone decreasing, which is the other condition of the alternating series test.) Therefore, the series above diverges and$x=-4$ is not in the set.

Substituting 0 for gives:

$\begin{array}{c}\sum _{n=0}^{\infty }\frac{{n}^{2}}{{2}^{n}}{\left(0+2\right)}^{n}=\sum _{n=0}^{\infty }\frac{{n}^{2}}{{2}^{n}}{2}^{n}\\ =\sum _{n=0}^{\infty }{n}^{2}\end{array}$

Here, use the fact that if a series converges, then $\underset{n\to \infty }{\mathrm{lim}}{a}_{n}=0$

By contraposition, it follows that

$p⇒q⇔¬q⇒¬p$

In this case, that is:

$\left(\text{Series converges}⇒\underset{n\to \infty }{\mathrm{lim}}{a}_{n}=0\right)⇔\left(\underset{n\to \infty }{\mathrm{lim}}{a}_{n}\ne 0⇒\text{Series diverges}\right)$

Now, for${a}_{n}={n}^{2}$it follows that

$\underset{n\to \infty }{\mathrm{lim}}{n}^{2}=\infty \ne 0$

Therefore, the series above diverges and$x=0$ is not in the set

.$x\in \left(-4,0\right)$