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Considering the following series

$$\sum_{n=1}^{\infty} \frac{n}{2^n}(x+2)^n + \sum_{n=1}^{\infty} \frac{n^3}{\sqrt {n!}}$$

We need to calculate the domain of convergence of this series.

Well the first series is a power series, it's easy to calculate the radius of convergence (it's 2). So the domain of convergence is $-4<x<0$

For the second series we need to see if the series in convergent or divergent, right? (Since it's a numerical series). Well I concluded it's divergent...

Now in $-4<x<0$ the series is divergent because it's the sum of a divergent series and a convergent series, right? But what happens outside the domain of convergence of the first series? Well, since we have two divergent series we can't conclude anything right? Then how should I proceed?. Thanks!

Granger Obliviate
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1 Answers1

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Note that for the second series, $\sum_{n=1}^\infty a_n$, where $a_n=\frac{n^3}{\sqrt{n!}}$, the ratio test asserts

$$\begin{align} \lim_{n\to \infty}\frac{a_{n+1}}{a_n}&=\lim_{n\to \infty}\left(\frac{(n+1)^3}{n^3}\frac{\sqrt{n!}}{\sqrt{(n+1)!}}\right)\\\\ &=\lim_{n\to \infty}\left(\left(1+\frac1n\right)^3\frac{1}{\sqrt{n+1}}\right)\\\\ &=0 \end{align}$$

Therefore, the series converges.

The sum of the two series converge for $-4<x<0$

Mark Viola
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