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 ...
 A: 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. 
A: It doesn't converge; your method is valid.  There is this thing called the Reimann Rearrangement theorem.  Consider$$\sum _1 ^\infty {(-1)^{x+1}\over x}$$  This gives ${1\over 1}-{1\over 2} +{1\over 3} -{1\over4}...$  It can be shown that this converges to $\ln2$.  However, notice that if you add up the positive terms, it goes to infinity, and the negative terms add up to negative infinity.  So you can actually rearrange the terms to get any number.  For example, let's say you wanted to get them to add to $\pi$.  You add only positive terms until they get to a value above $\pi$.  Then you subtract the first negative term, and get a value beneath $\pi$.  You add again till it goes above $\pi$, and then you subtract the next negative term, getting a value beneath $\pi$.  This process is repeated, and the value tends toward $\pi$.  This works for any series which is conditionally convergent.  The series you gave as an example is not.  You are allowed to rearrange them like that.  The method you used was correct.  Keep in mind the above, however, as that method may not always work.
