Does $\lim_{m \to \infty}\sum_{n=1}^m (-1)^n (\sum_{k=n^2}^{(n+1)^2-1}\frac{1}{\sqrt{k}}-2) $ exist? This question is based on
an answer and comment
to this question:
convergence of $\sum\limits_{n=1}^\infty \frac{(-1)^{\lfloor \sqrt{n}\rfloor}}{\sqrt{n}}$
Does
$\displaystyle \lim_{m \to \infty} \sum_{n=1}^m (-1)^n
\left[ \sum_{k=n^2}^{(n+1)^2-1}\frac{1}{\sqrt{k}}-2 \right]
$
exist?
The answers there show that
$\sum_{k=n^2}^{(n+1)^2-1}\frac{1}{\sqrt{k}}
\to 2
$,
but are not precise enough
to show that the difference is monotonic,
so the alternating series theorem
can not be applied.

We can express $2$ as the telescoping sum $$\sum_{k=n^2}^{(n+1)^2-1} (2\sqrt{k+1}-2\sqrt{k}),$$
which lets us rewrite the sum over $n$ as $$\sum_{n=1}^\infty (-1)^n \sum_{k=n^2}^{(n+1)^2-1} \left(\frac1{\sqrt k}-2\sqrt{k+1}+2\sqrt{k}\right).$$ 
I claim that this series is actually absolutely convergent, for which we just need to prove the convergence of $$\sum_{k=1}^\infty \left(\frac1{\sqrt k}-2\sqrt{k+1}+2\sqrt{k}\right).$$
Consider the inside term: $$\frac1{\sqrt k} - 2\sqrt{k+1} + 2\sqrt k = \frac{2k+1 - 2\sqrt{k(k+1)}}{\sqrt k}  = \frac{(\sqrt{k+1} - \sqrt k)^2}{\sqrt k}.$$
We have $$\left(\sqrt k + \frac1{2\sqrt k}\right)^2 = k + 1 + \frac1{4k} > k+1 \implies \sqrt k + \frac1{2\sqrt k} > \sqrt{k+1},$$
so $\sqrt{k+1} - \sqrt k < \frac1{2\sqrt k}$, and therefore $$\sum_{k=1}^\infty \left(\frac1{\sqrt k}-2\sqrt{k+1}+2\sqrt{k}\right) < \sum_{k=1}^\infty \frac1{4k \sqrt k}$$ which converges by the $p$-series test.
