Question

In problems 1-6, determine the convergence set of the given power series.

$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\frac{{\mathbf{n}}^{\mathbf{2}}}{{\mathbf{2}}^{\mathbf{n}}}{\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{2}\mathbf{)}}^{\mathbf{n}}$

Short answer

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

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,

$a_{n+1}=\frac{{(n+1)}^2}{2^{n+1}}$

and

$\begin{array}{c}\underset{n\to \infty }{\lim }\left|\frac{a_n}{a_{n+1}}\right|=\underset{n\to \infty }{\lim }\mid \frac{\frac{n^2}{2^n}}{\frac{{(n+1)}^2}{2^{n+1}}\mid }\\ =\underset{n\to \infty }{\lim }\left|\frac{n^2{2}^{n+1}}{2^n{(n+1)}^2}\right|\\ =\underset{n\to \infty }{\lim }|\frac{n^2}{{(n+1)}^2}\times \frac{2^{n+1}}{2^n}|\\ =\underset{n\to \infty }{\lim }|\frac{n^2}{{(n+1)}^2}\times 2^{n+1-n}|\\ =\underset{n\to \infty }{\lim }|\frac{n^2}{{(n+1)}^2}\times 2|\end{array}$

Simplify further.

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

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

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

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

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

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

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

(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 $\underset{n\to \infty }{\lim }{a}_n=0$

By contraposition, it follows that

$p\Rightarrow q\iff \neg q\Rightarrow \neg p$

In this case, that is:

$\left(\text{Seriesconverges}\Rightarrow \underset{n\to \infty }{\lim }{a}_n=0\right)\iff \left(\underset{n\to \infty }{\lim }{a}_n\ne 0\Rightarrow \text{Seriesdiverges}\right)$

Now, for$a_n=n^2$it follows that

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

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

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