$a_n=(-1)^{k_n}\frac{1}{n}$ where $(k_n-1)^2
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?