Is this series conditionally convergent or absolutely convergent? $ \sum_{k=1}^{\infty}\left(-1\right)^{k+1}\frac{k^{k}}{\left(k+1\right)^{k+1}} $ Is this series conditionally convergent or absolutely convergent? How to prove it ?

I tried ratio test and root test, but did not work I think. 

$
\sum_{k=1}^{\infty}\left(-1\right)^{k+1}\frac{k^{k}}{\left(k+1\right)^{k+1}}
$
To show that $\sum_{k=1}^{\infty}\left(-1\right)^{k+1}\frac{k^{k}}{\left(k+1\right)^{k+1}}$ converges, it suffices to show that $k \mapsto \frac{k^{k}}{\left(k+1\right)^{k+1}}$ is monotonically decreasing and then we can apply the alternating series test.

Consider $$f(x) = \ln\frac{x^x}{(x+1)^{x+1}} = x \ln x - (x+1)\ln (x+1)$$ We have $f'(x) = \ln x - \ln (x+1) &lt; 0$ so $f$ is monotonically decreasing.

Then $e^{f(x)} = \frac{x^x}{(x+1)^{x+1}}$ is also monotonically decreasing so the claim follows.
Hints

Use the alternating series test to determine whether the series is convergent. For absolute convergence use the limit comparison test, noting that
$$
\lim_{k\to \infty}\frac{k^k}{(k+1)^{k+1}}\biggr/\frac{1}{k}=\lim_{k\to \infty}
\left(1-\frac{1}{k+1}\right)^{k+1}=\frac{1}{e}
$$
and compare with the harmonic series.