TITLE: Is the series $\sum_{n=2}^\infty\frac{(-1)^{\lfloor\sqrt{n}\rfloor}}{\ln{n}}$ convergent?
QUESTION [7 upvotes]: Define
$$S_k=\sum_{n=2}^{k} \frac{(-1)^{\lfloor\sqrt{n}\rfloor}}{\ln{n}}$$.
Is $S_k$ or $S_{4k^2}$ convergent?
I think $S_k$ isn't convergent, since for big enough $n$, the $\lfloor\sqrt{n}\rfloor$ will change more slowly than for small $n$, but I don't know how to show it formally.
For $S_{4k^2}$, I think the series is also divergent since then $\lfloor\sqrt{n}\rfloor$ is always even, so we consider
$$S_{4k^2}=\sum_{n=2}^{k} \frac{1}{\ln{n}},$$
but we know $1/\ln n$ is divergent, so $S_{4k^2}$ is divergent. Is my reasoning here correct?

REPLY [3 votes]: Informally speaking, on the interval $n\in[A^2,A^2+2A]$ the value of $\frac{1}{\log n}$ is quite close to $\frac{1}{2\log A}$, hence we may expect that:
$$S_{M^2+2M}=\sum_{A=1}^{M}\sum_{n\in[A^2,A^2+2A]}\frac{(-1)^A}{\log n}\approx \sum_{A=1}^{M}\frac{(-1)^A (2A+1)}{2\log A}\tag{1}$$
but the general term of the last sum is not $o(1)$ as $A\to +\infty$. 
Since for any $n\in[A^2,A^2+2A]$ we have:
$$\frac{1}{\log A^2}-\frac{1}{\log n} = \frac{\log\frac{n}{A^2}}{\log n\log A^2}\leq\frac{\frac{2}{A}}{4\log^2 A}=\frac{1}{2A\log^2 A}$$
and
$$\sum_{A=1}^{+\infty}\frac{1}{A\log^2 A}<+\infty $$
by Cauchy's condensation test, the difference between the RHS and the LHS of $(1)$ is bounded. 
This proves that the original series is not convergent.