Does $1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{4}+\frac{1}{4^2}-\dots$ converge? 
Does
  $$1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{4}+\frac{1}{4^2}-\dots$$
  converge?

What I have tried: 
The series diverges since $$1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{4}+\frac{1}{4^2}-\dots=(1-\frac{1}{2})+(\frac{1}{2^2}-\frac{1}{3})+(\frac{1}{3^2}-\frac{1}{4})+\dots=\sum(\frac{1}{k^2}-\frac{1}{k+1})$$ and it can be shown that $$\sum \frac{1}{k+1}$$ diverges. 
However, I am not sure at all if this is correct or not. Am I allowed to re-group the terms like that? I think there would be a more elegant way to do this ...

You can prove directly that the series diverges by doing, for $k\geq3$,
$$
\frac1 {k^2}-\frac1 {k+1}=\frac {k+1-k^2}{(k+1)k^2}<\frac {-k^2}{2 (k+1)k^2}=-\frac12\,\frac1 {k+1}
$$
and the series diverges by comparison. 
