Convergence of the series $\sum \frac{(-1)^{\sqrt{n}}}{n}.$ I'm looking for some help to show that:
$$\sum {(-1)^{\lfloor \sqrt{n}\rfloor}\over n} < \infty$$
 A: After clarification, it seems that the goal is to prove that the sequence $(S_n)$ converges, where, for every $n\geqslant1$,
$$
S_n=\sum_{k=1}^n\frac{(-1)^{\lfloor k\rfloor}}k.
$$
To do so, consider, for every $n\geqslant1$,
$$
T_n=\sum_{k=n^2}^{(n+1)^2-1}\frac{(-1)^{\lfloor k\rfloor}}k=(-1)^n\sum_{k=n^2}^{(n+1)^2-1}\frac1k.
$$
For every $n$,
$$
|T_n|\leqslant\sum_{k=n^2}^{(n+1)^2-1}\frac1{n^2}=\frac{2n+1}{n^2}\leqslant\frac3n,
$$
hence $T_n\to0$. Furthermore, the signs of the entries $T_n$ alternate hence, if the sequence $|T_n|$ is noninceasing, the series
$$
\sum_{n\geqslant1}T_n
$$
is an alternating series and, as such, its sums converge to some limit $\ell$. For every $n$, there exists some $k$ such that $k^2\leqslant n\lt (k+1)^2$ hence
$$
\sum_{i=1}^kT_i\leqslant S_n\leqslant \sum_{i=1}^{k+1}T_i\quad\text{or}\quad\sum_{i=1}^{k+1}T_i\leqslant S_n\leqslant \sum_{i=1}^{k}T_i,
$$
depending on the parity of $k$. This proves that $S_n\to\ell$.
To conclude, it remains to show that indeed $|T_{n+1}|\leqslant|T_n|$ for every $n$. Can you do that?
