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

Question

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?

Thanks for reading

Answer 1

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.

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I hope this helps ^_^

Answer 2

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.

Answer 3

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 $0.97329$