Convergence test $\sum_{n=1}^{+\infty}(-1)^{\left\lfloor{\sqrt{n}}\right\rfloor}\frac{1}{\ln(n)}$ How can I try to check the convergence of series 
$$
\sum_{n=2}^{+\infty}(-1)^{\left\lfloor{\sqrt{n}}\right\rfloor}\frac{1}{\ln(n)}
$$
 A: Use Cauchy's convergence test and consider the following subsums
$$\sum_{n=k^2}^{k^2+2k} (-1)^{\lfloor \sqrt n \rfloor} \frac{1}{\ln(n)}$$
A: We can write the partial sums of the series of interest as
$$\begin{align}
\sum_{n=4}^{(N+1)^2-1} \frac{(-1)^{\lfloor \sqrt n\rfloor}}{\log(n)}&=\sum_{m=2}^N \sum_{n=m^2}^{(m+1)^2-1}\frac{(-1)^{\lfloor \sqrt n\rfloor}}{\log(n)}\\\\
&=\sum_{m=2}^N (-1)^m \sum_{n=m^2}^{(m+1)^2-1} \frac{1}{\log(n)}\tag1
\end{align}$$
It is sufficient to show that the sequence $S_m=\sum_{n=m^2}^{(m+1)^2-1} \frac{1}{\log(n)}$, on the right-hand side of $(1)$, is increasing (and hence $\lim_{m\to\infty}S_m\ne0$).  Proceeding we have the crude estimates
$$\begin{align}
S_m-S_{m-1}&=\sum_{n=m^2}^{(m+1)^2-1} \frac{1}{\log(n)}-\sum_{n=(m-1)^2}^{m^2-1} \frac{1}{\log(n)}\\\\
&\ge \frac{2m+1}{2\log(m+1)}-\frac{2m-1}{2\log(m-1)}\\\\
&=\frac{(m+1/2)\log(m-1)-(m-1/2)\log(m+1)}{\log(m+1)\log(m-1)}\tag2
\end{align}$$
We can estimate the numerator, $T_m$, of the expression on the right-hand side of $(2)$,  $T_m=(m+1/2)\log(m-1)-(m-1/2)\log(m+1)$ as
$$\begin{align}
T_m&=(m+1/2)\log(m-1)-(m-1/2)\log(m+1)&\\\\
&=\log(m)+m\log\left(1-\frac2{m+1}\right)+\frac12\log\left(1-\frac1{m^2}\right)\\\\
&\ge \log(m)-\frac{2m}{m-1}-\frac1{m^2}\tag3
\end{align}$$
From $(3)$, $T_m>0$ whenever $m$ is sufficiently large ($m>9$)$.
And we are done!
