convergence of $\sum_{n=1}^{\infty }\frac{(-1)^{[\sqrt{n}]}}{n}$ where $[a]$ means floor function of $a$

Question

I have a problem with solving the convergence of $\sum_{n=1}^{\infty }\frac{(-1)^{[\sqrt{n}]}}{n}$ where $[a]$ means floor function of $a$. I've already made the first thep and rewrote it with the $\sum_{n=1}^{\infty }(-1)^{n}\sum_{k=n^2}^{(n+1)^2-1}\frac{1}{k}$, which converges if and only if the previous one converges. Now I want to use Leibniz criterion, but I cannot prove that the second sum in the previous expression is monotonic. Does anyone know how to do it? (I cannot use Riemann integral, we haven't defined it yet)

Answer 1

Let us recall the definition of the hamronic numbers $$\sum_{k=1}^{N}\frac{1}{k}=H_{N} $$ hence $$S=\sum_{n\geq1}\frac{\left(-1\right)^{\left\lfloor \sqrt{n}\right\rfloor }}{n}=-1+\sum_{n\geq2}\frac{\left(-1\right)^{\left\lfloor \sqrt{n}\right\rfloor }}{n} $$ $$=-1+\sum_{n\geq2}\left(-1\right)^{n}\left(H_{\left(n+1\right)^{2}-1}-H_{n^{2}-1}\right) $$and using $$H_{n}=\log\left(n\right)+\gamma+O\left(\frac{1}{n}\right) $$ we get $$S=-1+\sum_{n\geq2}\left(-1\right)^{n}\log\left(\frac{\left(n+1\right)^{2}-1}{n^{2}-1}\right)+O\left(1\right) $$ $$=-1+\sum_{n\geq2}\left(-1\right)^{n}\log\left(\frac{n\left(n+2\right)}{n^{2}-1}\right)+O\left(1\right)$$ and so we have the convergence for the Leibniz' criterion.