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?
 A: It may help to rearrange the proposed general term as follows:
\begin{align*}
\sqrt{n + 1} - \sqrt{n} = \frac{(n + 1) - n}{\sqrt{n + 1} + \sqrt{n}} = \frac{1}{\sqrt{n + 1} + \sqrt{n}}
\end{align*}
which is a decreasing sequence of non-negative terms.
Finally, due the Leibniz test, the proposed alternating series converges.
Hopefully this helps!
A: A slightly more general technique (though the one by Átila Correia is the simplest and probably best here):

Suppose you have a (positive) sequence $(a_n)_n$, not necessarily monotone, such that
$$
a_n = b_n + c_n + O(c_n)
$$
where $(b_n)_n$ is non-increasing with limit $0$, and $(c_n)_n$ is absolutely convergent. Then $\sum_n (-1)^n a_n$ converges.

Indeed, you have
$$
\sum_{n=1}^N (-1)^n a_n = \sum_{n=1}^N (-1)^n b_n + \sum_{n=1}^N (-1)^n (c_n+O(c_n))
$$
and on the RHS the first term will converge (by Leibniz's criterion) and the second will converge (absolutely) (by comparison with $\sum_n c_n$).
Now, in your case, $$\begin{align}a_n &= \sqrt{n+1}-\sqrt{n}
= \sqrt{n}\left(\sqrt{1+\frac{1}{n}}-1\right) = \sqrt{n}\left(\frac{1}{2n}-\frac{1}{8n^2} + O\!\left(\frac{1}{n^2}\right)\right) \\&= \frac{1}{2\sqrt{n}}-\frac{1}{8n^{3/2}} + O\!\left(\frac{1}{n^{3/2}}\right)\end{align}$$
and you can apply the above with $b_n = \frac{1}{2\sqrt{n}}$, $c_n = -\frac{1}{8n^{3/2}}$.
A: 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.
A: Does it help to rearrange by summarizing every two terms:
$$\sum_{n = 1}^\infty (-1)^n (\sqrt{n + 1} - \sqrt{n}) = \sum_{k = 1}^\infty \sqrt{k + 2} - \sqrt{k + 1} - \sqrt{k + 1} + \sqrt{k}$$
and using a telescope trick?
