1
$\begingroup$

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!

$\endgroup$
  • 1
    $\begingroup$ The second series is convergent since for big enough $n$ we have $n! > n^{10}$. $\endgroup$ – user9077 Jun 4 '16 at 15:41
1
$\begingroup$

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 $-4<x<0$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.