Is this series convergent? $a_n=(-1)^{k_n}\frac{1}{n}$ where $(k_n-1)^2<n\leq k_n^2$. Is the series $\sum_n a_n$ convergent? I tried with all the classical methods, but they seem to fail, any hint?
EDIT: I had an idea: $\sum a_n=\sum_k (-1)^kb_k$ where $b_k=\sum_{n=(k-1)^2+1}^{k^2}\frac{1}{n}$, I proved that $b_k\rightarrow0$ I want to prove that it's decreasing, so I can apply Leibniz. So I want to prove that $b_{k+1}\leq b_k$, this is false for $k=1$, but it's true for $k=2,3,4$, so I suppose that the sequence of the $b_k$'s is decreasing if $k>1$, but I need some help to prove it.
EDIT EDIT: I checked with my pc and it's true that $b_{k+1}\leq b_k$ for $k>1$, but I still don't know how to prove it, any hint?
 A: Write it as follows (suppressing superfluous symbols). Note that squares are doubly used as indices, so I'm adding a compensatory $\mathcal{O}(1)$ into the fray to keep things formal:
$$\mathcal{O}(1)+\left(\sum_1^4-\sum_4^9\right)+\left(\sum_9^{16}-\sum_{16}^{25}\right)+\cdots +\left(\sum_{(2m-1)^2}^{(2m)^2}-\sum_{(2m)^2}^{(2m+1)^2}\right)+\cdots$$
Now use the asymptotic $H_n \sim \log n+\gamma+\frac{1}{2n}+\mathcal{O}(n^{-2})$ on the individual terms:
$$\sum_{(2m-1)^2}^{(2m)^2}-\sum_{(2m)^2}^{(2m+1)^2}=2\log\left(\frac{2m}{2m-1}\right)-2\log\left(\frac{2m+1}{2m}\right) +\mathcal{O}\left(\frac{1}{m^2}\right).$$
Add into the mix $\log(1+x)=x+\mathcal{O}(x^2)=\log\left(\frac{1}{1-x}\right)$ and we have $\square =\mathcal{O}(m^{-2})$, which shows that this series converges. This may or may not help you, depending on what you can use in your HW.
A: Write 
$$b_k=\sum\limits_{n=1}^{2k-1}\frac1{(k-1)^2+n},\qquad
b_{k+1}=c_k+\sum\limits_{n=1}^{2k-1}\frac1{k^2+n},
$$
with
$$
c_k=\frac1{k^2+2k}+\frac1{k^2+2k+1}. 
$$
Then $b_k\gt b_{k+1}$ if and only if $d_k\gt c_k$ with
$$
d_k=\sum\limits_{n=1}^{2k-1}\left(\frac1{(k-1)^2+n}-\frac1{k^2+n}\right)=
\sum\limits_{n=1}^{2k-1}\frac{2k-1}{(k^2+n)((k-1)^2+n)}.
$$
For every $1\leqslant n\leqslant 2k-1$, $k^2+n\leqslant k^2+2k$ and $(k-1)^2+n\leqslant k^2$ hence
$$
d_k\geqslant\frac{(2k-1)^2}{k^3(k+2)}.
$$
On the other hand $k^2+2k+1\geqslant k^2+2k$ hence
$$
c_k\leqslant\frac2{k(k+2)}.
$$
One sees that $c_k\lt d_k$ as soon as
$$
\frac2{k(k+2)}\lt\frac{(2k-1)^2}{k^3(k+2)},
$$
that is,
$2k^2\lt(2k-1)^2$, which is equivalent to $2k^2-4k+1\gt0$, which holds for every $k\geqslant2$. Thus, $b_k\gt b_{k+1}$ for every $k\geqslant2$.
