Prove that the infinite series is divergent I would like to know how I proof that the following infinite series is divergent:
$$\sum_{i=1}^n (-1)^n*(\frac{n}{n+1})^n$$
I tried the following tests: ratio test, root test, Leibniz criterion, divergence test, but they lead to an inconculsive result. The other only tests I am allowed to use is the comparison test and Cauchy's convergence test , but I don't know how to proof that the infinite series is divergent with these.
The series has the $n^{th}$ term $a_n=\dfrac{1}{\left(-1-\dfrac{1}{n}\right)^n}$ does not converge to $0$. Thus the series diverges.
Observe
\begin{align}
\limsup_{n\rightarrow \infty}\ (-1)^n \left(\frac{n}{n+1}\right)^n = e^{-1} \neq 0.
\end{align}

Edit: By the request of Ted Shifrin, I have updated my answer. 

Heuristically speaking, an infinite series converges provided the terms that you are adding eventually becomes negligible. Otherwise, if you keep on adding ''big" terms to your partial sum
\begin{align}
\sum^N_{n=1} a_n + "\text{infinitely many big terms}"
\end{align}
then there's no hope that the series will converge. For instance consider the example
\begin{align}
-1 + 1 + -1 + 1 + -1+\ldots 
\end{align}
We see that if we add an even number of terms then the partial sum  will be $0$, but if we add an odd number of terms the partial sum will be $-1$. Hence our partial sum will fluctuate between $0$ and $-1$, which means the sequence of partial sums will not converge. The reason why the partial sum does not converge is mainly due to that fact that each term we add to the  partial sum is not negligible. In fact the new term that we added to the partial sum greatly affects the value of the partial sum. 

Back to the problem. In the given series, we see that for large enough $N$ the series looks more or less like the following
\begin{align}
\ldots + e^{-1}-e^{-1} + e^{-1}-e^{-1} + \ldots 
\end{align}
which is almost exactly the same as our above example. Hence there's no way the given series converges.