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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.
