Determine whether series converges Given this series:
$\sum\limits_{n=1}^{\infty}\frac{(n+1)^{n-1}}{(-n)^n}$
What I have tried is root test but ending up with something like $ \sqrt[\leftroot{10} n]{(n+1)^{n-1}} $
Ratio test is similiar -> failing
Leibniz:
but i cant help myself with monotonicity proof (by induction?)
$\sum\limits_{n=1}^{\infty}\frac{(n+1)^{n-1}}{(-n)^n} \iff\sum\limits_{n=1}^{\infty} (-1)^n \frac{(n+1)^{n-1}}{n^n}$ <- is that correct?
Then proof that $n^n > (n+1)^{n-1}$
And proof that $(n+1)^{n+1}> (n+2)^n$
But I failed doing so.
Pls give me a hint...
Thank you in advance!
 A: Let the given series be $\,\displaystyle\color{brown}{S_{\infty}=\sum^{\infty}_{n=1}{(-1)}^{n}\,{x}_{n}}\,\,$ , where $\,\color{crimson}{{x}_{n}=\dfrac{{(n+1)}^{n-1}}{{n}^{n}}}\,$
Now, consider $$\displaystyle\dfrac{{x}_{n+1}}{{x}_{n}}=\dfrac{{(n+2)}^{n}}{{(n+1)}^{n+1}}\cdot\dfrac{{n}^{n}}{{(n+1)}^{n-1}}$$
$$\displaystyle\implies\dfrac{{x}_{n+1}}{{x}_{n}}=\dfrac{\left({n}^{2}+2n\right)^{n}}{{(n+1)}^{2n}}$$
$$\displaystyle\implies\dfrac{{x}_{n+1}}{{x}_{n}}=\dfrac{\left({n}^{2}+2n\right)^{n}}{\left({n}^{2}+2n+1\right)^{n}}<1$$
$$\displaystyle\implies\dfrac{{x}_{n+1}}{{x}_{n}}<1$$
$$\displaystyle\implies{x}_{n+1}<{x}_{n}$$
So, the sequence $\{x_{n}\}$ is monotone decreasing.

According to Leibnitz's test for alternating series,
If $\,\color{magenta}{\{{x}_{n}\}}\,$ is a monotone decreasing sequence of positive real numbers and $\,\displaystyle\color{magenta}{\lim_{n\to\infty}\,{x}_{n}=0}\,$ , then the alternating series $\,\displaystyle\color{magenta}{\sum^{\infty}_{n=0}{(-1)}^{n}{x}_{n}}\,$ is convergent.

Now, let us find the limit
$$\displaystyle\lim_{n\to\infty}\,{x}_{n}$$
$$\displaystyle=\lim_{n\to\infty}\,\dfrac{{(n+1)}^{n}}{{n}^{n}}\cdot{(n+1)}^{-1}$$
$$\displaystyle=\lim_{n\to\infty}\left(1+\dfrac{1}{n}\right)^{n}\cdot\lim_{n\to\infty}\dfrac{1}{n+1}$$
$$\displaystyle\lim_{n\to\infty}\,{x}_{n}=e\cdot0$$
$$\displaystyle\lim_{n\to\infty}\,{x}_{n}=0$$
Hence, the given series is convergent
