# I am having difficulty finding if this series converges or d

I am having difficulty finding if this series converges or diverges:
$\sum _{n=0}^{\mathrm{\infty }}\frac{{\left(-2\right)}^{n+1}}{{n}^{n-1}}$
I am unsure of which test to use. At first, I thought I should use alternating series test, but I am unable to manipulate the numerator to ${\left(-1\right)}^{n+1}$. I thought I may be able to use root test, but I have no clue how to manipulate the series to do that.
What series test would I use, and how?
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Ethan Sanders
Alternating series test
Note that
$\sum _{n=0}^{\mathrm{\infty }}\frac{{\left(-2\right)}^{n+1}}{{n}^{n-1}}=\sum _{n=0}^{\mathrm{\infty }}{\left(-1\right)}^{n+1}\frac{{2}^{n+1}}{{n}^{n-1}}$
By the alternating series test this converges if
1. ${a}_{n}>0$: we can easily this is satisfied
2. ${a}_{n}$ is monotic: ${a}_{n}$ is decreasing monotonically
3. $\underset{n\to \mathrm{\infty }}{lim}{a}_{n}=0$ I will leave this one to you.
Comparison test
$|\sum _{n=0}^{\mathrm{\infty }}\frac{{\left(-2\right)}^{n+1}}{{n}^{n-1}}|\le \sum _{n=0}^{\mathrm{\infty }}\frac{{2}^{n+1}}{{n}^{n-1}}\le \sum _{n=0}^{3}{\left(\frac{2}{n}\right)}^{n+1}+\sum _{n=4}^{\mathrm{\infty }}{\left(\frac{2}{n}\right)}^{n+1}$
The last series is a geometric series and therefore converges.
###### Not exactly what you’re looking for?
Virginia Palmer
Hint: what can you say about $\frac{{2}^{n-1}}{{n}^{n-1}}$?
Can you show that $\frac{{2}^{n-1}}{{n}^{n-1}}\le \frac{1}{{n}^{2}}$ for $n\ge N$?
###### Not exactly what you’re looking for?
user_27qwe

Making the problem more general, let

$\mathrm{log}\left({a}_{n}\right)=\left(n+1\right)\mathrm{log}\left(x\right)-\left(n-1\right)\mathrm{log}\left(n\right)$
$\mathrm{log}\left({a}_{n+1}\right)-\mathrm{log}\left({a}_{n}\right)=\left(n-1\right)\mathrm{log}\left(n\right)-n\mathrm{log}\left(n+1\right)+\mathrm{log}\left(x\right)$
Using Taylor for large values of n
$\mathrm{log}\left({a}_{n+1}\right)-\mathrm{log}\left({a}_{n}\right)=\mathrm{log}\left(\frac{x}{ne}\right)+\frac{1}{2n}+O\left(\frac{1}{{n}^{2}}\right)$
$\frac{{a}_{n+1}}{{a}_{n}}={e}^{\mathrm{log}\left({a}_{n+1}\right)-\mathrm{log}\left({a}_{n}\right)}\sim \frac{x}{ne}\to 0$