Does $\sum_{k=1}^\infty(-1)^k(\frac{k}{k+1})^k$ converge? 
Let $$\sum_{k=1}^\infty(-1)^k(\frac{k}{k+1})^k.$$Does it converge? Does it converge absolutely?

Now $$\lim_{k\to\infty}|(-1)^k(\frac{k}{k+1})^k|=...=\frac{1}{e}\neq0.$$ Thus the series isn't absolutely convergent. I cannot apply the Leibniz theorem, because $\lim_{k\to\infty}(\frac{k}{k+1})^k=\frac{1}{e}\neq0.$ 

Hint:
Does
$$\lim_{k\to\infty}(-1)^k\left(\frac{k+1}k\right)^k=0\;\;?$$
