Proof that $\sum (-1)^n a_n$ converge with $a_n = (\sqrt{n+2}-\sqrt{n+1})\sin\frac 1n$

Welcome to MSE!

Informally, to show that a sum converges $\sum a_n$, we want to show that the $a_n$ go to $0$ quickly.

So when I see this problem, the first thing I notice is the $\sin \left ( \frac{1}{n} \right )$ term, which we can approximate as the (much simpler) $\frac{1}{n}$. After all, $\sin(x) \leq x$ and this approximation is very good for $x \approx 0$.

Then we need to handle the $\left ( \sqrt{n+2} - \sqrt{n+1} \right )$ term. Intuitively, this should also go to $0$, since for large $n$, $n+2 \approx n+1$.

A common trick to make this intuition precise is to use the mean value theorem. It tells us that

$$ \frac{\sqrt{n+2} - \sqrt{n+1}}{1} = \frac{\sqrt{n+2} - \sqrt{n+1}}{(n+2) - (n+1)} = \frac{1}{2 \sqrt{\xi}} $$

for some $n+1 \leq \xi \leq n+2$. (Since the derivative of $\sqrt{x}$ is $\frac{1}{2 \sqrt{x}}$.)

Now we can upper bound this sum, since $\frac{1}{2 \sqrt{\xi}} \leq \frac{1}{2 \sqrt{n}}$ when $n \leq \xi$, and taken together we find

$$a_n = \left ( \sqrt{n+2} - \sqrt{n+1} \right ) \sin \left ( \frac{1}{n} \right ) \leq \frac{1}{2 \sqrt{n}} \frac{1}{n}$$

Of course, then $\sum a_n \leq \frac{1}{2} \sum \frac{1}{n^{1.5}}$, which converges by comparison with a p-series.

Notice this makes precise the intuition that we had at the beginning: We showed that the $a_n \to 0$ "quickly" (by which we mean "like $\frac{1}{n^{1.5}}$"), which is enough to get convergence.

I hope this helps ^_^