Does $\sum\frac{(-1)^{\lfloor \sqrt n\rfloor}}{ n}$ converge? Does $\sum\frac{(-1)^{\lfloor \sqrt n\rfloor}}{n}$ converge ?
I know there is a technical method that uses an adapted Abel's rule 
$(*)$ ; I would like to know if you see another method (which is understable with basic knowledge) to prove it converges. Thank you.
Remark: Actually, it can be proved that for all $z \in \mathbb C$ such that $|z| = 1$, $\sum\frac{(-1)^{\lfloor \sqrt n\rfloor}}{ n}z^n$ converges.
$(*)$ Consider two series $\sum a_n$ and $\sum b_n$. Define $B_m = \sum_{n=0}^mb_n$ 
If the sequence $(a_n)$ tends to $0$ and the sequence $(B_n)$ is bonded, and if the series $\sum (a_{n+1} - a_n)$ converges absolutely, then $\sum a_nb_n$ converges.
You have to adapt this theorem to use it here.
 A: Hint:
The sign is constant from $k^2$ to $(k+1)^2$, asymptotically corresponding to terms
$$\int_{k^2}^{(k+1)^2}\frac{dx}x=\log\left(1+ \frac1k\right)^2\approx \frac2k$$
and the alternating sum of $\dfrac1k$ is known to converge (to $\log2$).
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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Hint:
We can find a pattern:
\begin{align}
&\bbox{\ds{\sum_{n = 1}^{\infty}{\pars{-1}^{\left\lfloor{\root{n}}\right\rfloor} \over n}}}
\\[5mm] = &\
-\pars{{1 \over 1} + {1 \over 2} + {1 \over 3}} +
\pars{{1 \over 4} + {1 \over 5} + {1 \over 6} +{1 \over 7} + {1 \over 8}}
\\[2mm] &\
-\pars{{1 \over 9} + {1 \over 10} + {1 \over 11} + {1 \over 12} + {1 \over 13} +
 + {1 \over 14} + {1 \over 15}}
\\[2mm] &
+\pars{{1 \over 16} + {1 \over 17} + {1 \over 18} + {1 \over 19} + {1 \over 20} 
 + {1 \over 21} + {1 \over 22} + {1 \over 23} + {1 \over 24}}
\\[2mm] &
-\pars{{1 \over 25} + {1 \over 26} + {1 \over 27} + {1 \over 28} + {1 \over 29} 
 + {1 \over 30} + {1 \over 31} + {1 \over 32} + {1 \over 33} + {1 \over 34} + {1 \over 35}} + \cdots
\\[1cm] = &\
-H_{3} + \pars{H_{8} - H_{3}} - \pars{H_{15} - H_{8}} + \pars{H_{24} - H_{15}} -
\pars{H_{35} - H_{24}} + \cdots
\end{align}

Note that each group in $\ds{\pars{}}$'s
  $\ds{\pars{~\mbox{namely,}\ \pars{-1}^{n + 1}\pars{H_{n^{2} + 4n + 3} -
H_{n^{2} + 2n}}\,,\ n = 1,2,3,\ldots}~}$ is $\ds{\stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}}$ to
  $\ds{\pars{-1}^{n + 1}\,{2 \over n}}$ which agrees with
  $\texttt{@Ives Daoust}$ Answer.

