Does the series $\sum_{n=1}^{\infty} (-1)^n(\sqrt{n+1}-\sqrt{n})$ converge or diverge? $$\sum_{n=1}^{\infty} (-1)^n(\sqrt{n+1}-\sqrt{n})$$
I know that $\sum_{n=1}^{\infty} \sqrt{n+1}-\sqrt{n}$ diverges since the kth partial sum of the telescoping series can be written as:
$(\sqrt{2}-\sqrt{1})+(\sqrt{3}-\sqrt{2})+(\sqrt{4}-\sqrt{3})+...+(\sqrt{k+1}-\sqrt{k})=\sqrt{k+1}-1$
Then taking the limit as $k\rightarrow\infty$ of the kth partial sum gives me
$\lim_{n\rightarrow\infty}=\sqrt{k+1}-1=\infty$
So this series is not absolutely convergent. But how can I check if it is conditionally convergent?
Intuitively I'm thinking if $\sqrt{n+1}-\sqrt{n}$ diverges then
$\sum_{n=1}^{\infty} (-1)^n(\sqrt{n+1}-\sqrt{n})$ must diverge too. Is my logic correct here?
The solution says that this series is conditionally convergent but I'm not sure how they get that? Do I need to use the Alternating series test?

Setting $\displaystyle a_{n}:=(\sqrt{n+1}-\sqrt{n})\cdot\left(\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}\right)=\frac{1}{\sqrt{n+1}+\sqrt{n}}$ so $\displaystyle \sum_{n=1}^{+\infty}(-1)^{n}a_n$ converges because $(a_n)_{n\in \Bbb{N}}$ is positive terms, decreasing (because the "denominator" increasing so "1/denominator" decreasing) and converges to zero and $(-1)^n$ is uniformly bounded by $1$ respect to $n$. However as you pointed out $\displaystyle \sum_{n=1}^{+\infty}\left|(-1)^n a_n\right|$ does not converges (alternatively for your solution in this part you can see $\displaystyle \sum_{n=1}^{+\infty}|(-1)^{n}a_{n}|=\sum_{n=1}^{+\infty}a_{n} \geqslant \frac{1}{2}\sum_{n=1}^{+\infty}\frac{1}{\sqrt{n+1}}$ diverges) . So the series $\displaystyle 
 \sum_{n=1}^{+\infty}(-1)^{n}\left(\sqrt{n+1}-\sqrt{n}\right)$ is conditionally convergent.
