Does $\sum_{k=1}^n\frac{(-1)^k}{\sqrt{k}}$ converge?
My attempt:
$$\begin{aligned} \sum_{k=1}^{2n}\frac{(-1)^k}{\sqrt{k}}&= \sum_{k=1, 3, ..., 2n-1}\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\right)\\ &=\sum_{k=1, 3,..., 2n-1}\frac{1}{\sqrt{k(k+1)}(\sqrt{k+1}+\sqrt{k})} \end{aligned}$$ $$\begin{aligned} |S_{2n}-S_{2m}|&=\sum_{k=2m+1, 2m+3,..., 2n-1}\frac{1}{\sqrt{k(k+1)}(\sqrt{k+1}+\sqrt{k})}\\ &<\sum_{k=2m+1, 2m+3, ..., 2n-1}\frac{1}{2k\sqrt{k}}\\ &< ? \end{aligned}$$
Yes, it converges, by the alternating series test.
On the other hand, grouping the terms of your series in pairs will not help you to prove that it converges. The series$$1-1+1-1+1-1+\cdots$$diverges, but the series$$(1-1)+(1-1)+(1-1)+\cdots$$converges.
Your idea is fine but we need to consider
$$ \sum_{k=1}^{2n+1}\frac{(-1)^k}{\sqrt{k}} =-1+ \sum_{k=2}^{2n+1}\frac{(-1)^k}{\sqrt{k}}=-1+\sum_{k=1}^n\left(\frac{1}{\sqrt{2k}}-\frac{1}{\sqrt{2k+1}}\right) $$
and since
$$\frac{1}{\sqrt{2k}}-\frac{1}{\sqrt{2k+1}}=\frac{\sqrt{2k+1}-\sqrt{2k}}{\sqrt{2k}\sqrt{2k+1}}=\frac{1}{\sqrt{2k}\sqrt{2k+1}(\sqrt{2k+1}+\sqrt{2k})}\le\frac{1}{4k\sqrt{2k}}$$
we can conclude that the given series converges also by p-test.
You can use your idea to conclude. The subsequence $(S_{2n})_n$ of even index partial sums of the series $$ S_{2n}=\sum_{k=1}^{2n}\frac{(-1)^k}{\sqrt{k}} =\sum_{k=1}^n\underbrace{\frac{1}{\sqrt{k(k+1)}(\sqrt{k+1}+\sqrt{k})}}_{\displaystyle =O\Big(\frac{1}{k^{3/2}}\Big)} $$ converges to some limit $\ell$ since the RHS converges absolutely. Since the general term $\frac{(-1)^k}{\sqrt{k}}$ of the series tends to zero, subsequence $(S_{2n+1})_n$ of odd index partial sums of the series converges as well and to the same limit: $$S_{2n+1}=\underbrace{S_{2n}}_{\to\ell}-\underbrace{\frac{1}{\sqrt{2n+1}}}_{\to 0}\longrightarrow\ell$$ A sequence $(u_n)$ whose subsequences of even index terms $(u_{2n})_n$ and odd index terms $(u_{2n+1})_n$ both converge and have the same limit is itself convergent (to that same limit) and so the series converges.
