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

I've a problem about proving that a certain series converge and truly need some help.

The question is :

Let $$a_n = \left ( \sqrt{n+2}-\sqrt{n+1} \right ) \sin\frac 1n$$, what can you say about $$\sum a_n$$ The answer I need to get right is that both $$\sum a_n$$ and $$\sum (-1)^n a_n$$ are convergent.

I have two main questions about it :

1. I think I know how to prove that $$\sum a_n$$ is convergent but I find it highly slow : Basically I simplify $$a_n = \left ( \sqrt{n+2}-\sqrt{n+1} \right ) \sin \frac{1}{n} = \frac{\sin(\frac{1}{n})}{\sqrt{n+2}+\sqrt{n+1}}$$ and then I use the theorem that goes like if $$a_n$$ and $$b_n$$ are strictly superior to $$0$$ and that if $$\alpha = \lim_{n\to\inf} \frac{a_n}{b_n} > 0$$ and that the limit exist, then either both $$\sum b_n$$ and $$\sum a_n$$ are convergent or either they are both divergent. As $$b_n$$ I use $$\frac{\frac{1}{n}}{\sqrt{n+2}+\sqrt{n+1}}$$ that way alpha is equal to one and $$b_n$$ can be compared to $$\sum \frac{1}{n^{3/2}}$$ that converge. So now I know that $$\sum a_n$$ is convergent but I can't use the same trick with the $$(-1)^n$$ series since they would sometimes be under $$0$$ Am I in a too complicated way ?

2. I know that series that are absolutely convergent are convergent as well but the other way is not necessarily true, and I always struggle to prove that a $$\sum (-1)^n a_n$$ is convergent even when I know that $$\sum a_n$$ is... Am I missing a simple theorem?

• This is an alternating series. The only thing you need to check is that $a_n\to 0$. For this use the fact that $\sin 1/n\sim 1/n$. Jan 10, 2022 at 21:33
• You also need that it's monotonically decreasing in absolute value. Consider $\sum (\frac 1n - \frac {1}{2n})$, which is alternating with terms that go to $0$, but diverges. Jan 10, 2022 at 21:43
• Yeah definitely much more simpler that way thank you ! Jan 10, 2022 at 21:52
• When a series is absolutely convergent, you can change the signs arbitrarily without losing convergence.
– user1010241
Jan 10, 2022 at 22:12
• If $\sum a_n$ is convergent, and the $a_n$ are positive, then $\sum (-1)^n a_n$ is also convergent. You don't need the terms to decrease monotonically. Jan 10, 2022 at 22:12

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 ^_^

• For the square root term, I find it easier to multiply and divide by the conjugate: $$\sqrt{n+2}-\sqrt{n+1}=\frac {n+2-(n+1)}{\sqrt{n+2}+\sqrt{n+1}}= \frac{1}{\sqrt{n+2}+\sqrt{n+1}} \lt \frac{1}{2 \sqrt n}.$$ Jan 10, 2022 at 21:47
• Thanks it helped a lot, it's way easier that way and seems less prone to mistakes ! Jan 10, 2022 at 21:54

From your initial work, notice that we can make numerators larger and denominators smaller to get a fraction that's larger. So for positive integers $$n$$,

$$0\le \frac{\sin(\frac{1}{n})}{\sqrt{n+2}+\sqrt{n+1}} \le \dfrac{1/n}{\sqrt{n}+\sqrt{n}} = \dfrac{1}{2n^{3/2}}$$

Then the comparison test finishes showing the series is absolutely convergent.

• Well that's very simple like that thanks you, definitely the quickest method I think Jan 11, 2022 at 14:29

Using Taylor expansion for large values of $$n$$, you could show that $$\frac 1{2 n^{3/2}}-\frac 3{8 n^{5/2}}<\Big[\sqrt{n+2}-\sqrt{n+1}\Big] \sin \left(\frac{1}{n}\right)<\frac 1{2 n^{3/2}}$$ and then $$\frac{1}{2}\zeta \left(\frac{3}{2}\right)-\frac{3}{8}\zeta \left(\frac{5}{2}\right)<\sum_{n=1}^\infty\Big[\sqrt{n+2}-\sqrt{n+1}\Big] \sin \left(\frac{1}{n}\right)<\frac{1}{2}\zeta \left(\frac{3}{2}\right)$$ Numerically, $$0.80313<\sum_{n=1}^\infty\Big[\sqrt{n+2}-\sqrt{n+1}\Big] \sin \left(\frac{1}{n}\right)<1.30619$$ while the summation is $$0.97329$$

• So many different methods I did not see wow thank you ! Jan 11, 2022 at 14:30