# Prove convergence / divergence of $\sum_{n=2}^\infty(-1)^n\frac {\sqrt n}{(-1)^n+\sqrt n}\sin\left(\frac {1}{\sqrt n}\right)$

How can I prove or disprove the following series converges?

$$\sum_{n=2}^\infty(-1)^n\cfrac {\sqrt n}{(-1)^n+\sqrt n}\sin\left(\frac {1}{\sqrt n}\right)$$

I tried several things, none of which worked.

I wanted to use Abel's test or Dirichlet's test. I know that $\sin(\frac {1}{\sqrt n})$ is monotonically decreasing to $0$, but I wasn't able to show that $\Sigma_{n=2}^\infty(-1)^n\frac {\sqrt n}{(-1)^n+\sqrt n}$ is convergent, as it does not converge absolutely since $\frac {\sqrt n}{(-1)^n+\sqrt n}$ converges to 1. Neither was I able to show that the partial sum sequence of $\frac {\sqrt n}{(-1)^n+\sqrt n}$ is bounded. I'm at a loss. Would love any help.

Note - This exact question was discussed here a few years ago, but was not answered then and the hint provided in the responses was not useful.

• Use the fact that $\lim\limits_{x\rightarrow 0} \frac{\sin{x}}{x}=1$ Dec 16, 2017 at 10:51

Observe that $$\frac{\sqrt{n}}{\sqrt{n} + (-1)^n} = \frac{\sqrt{n}}{\sqrt{n} + (-1)^n} \cdot \frac{\sqrt{n} - (-1)^n}{\sqrt{n} - (-1)^n} = \frac{n - (-1)^n \sqrt{n}}{n-1},$$ hence the general term $a_n$ of your series can be written as $$a_n = b_n + c_n, \qquad b_n := (-1)^n \frac{n}{n-1} \sin \frac{1}{\sqrt{n}}, \quad c_n := - \frac{\sqrt{n}}{n-1} \sin \frac{1}{\sqrt{n}}\,.$$ Now, $\sum_n b_n$ is convergent by Leibnitz's criterion for alternating series, whereas $\sum c_n$ diverges to $-\infty$ (since $c_n \sim - 1/n$). So $\sum a_n$ diverges to $-\infty$.

• This is very nice and elegant. Thank you very much!
– Noa
Dec 16, 2017 at 11:13
• I have been thinking about this for a few days and I came to think that this solution is unfortunately incorrect, because it involves rearranging the terms and then summing them in a different order. By Riemann's rearrangement theorem, we've actually done nothing...
– Noa
Dec 19, 2017 at 17:21
• There is no rearrangement, since, for every $n$, $a_n = b_n + c_n$. So you can reason on partial sums: if $A_n := \sum_{k=1}^n a_k$, and $B_n$, $C_n$ are defined in a similar manner, then $A_n = B_n + C_n$. Taking the limit as $n\to +\infty$, you get $\lim_n A_n = -\infty$. Dec 19, 2017 at 20:21
• Got it. Thanks for your help, I really appreciate it!
– Noa
Dec 19, 2017 at 20:55

Hint. By Taylor series expansions, one has, as $x \to 0$, $$\sin x=x+O(x^3), \qquad \frac1{1+x}=1-x+O(x^2),$$giving, as $n \to \infty$, $$(-1)^n\frac {\sqrt n}{(-1)^n+\sqrt n}=\frac {(-1)^n}{1+\frac{(-1)^n}{\sqrt n}}=(-1)^n-\frac{1}{\sqrt n}+O\left(\frac {1}{n} \right),$$and$$\sin\frac {1}{\sqrt n}=\frac {1}{\sqrt n}+O\left(\frac {1}{n^{3/2}} \right).$$ Then, as $n \to \infty$, one gets

$$(-1)^n\frac {\sqrt n}{(-1)^n+\sqrt n}\:\sin \frac {1}{\sqrt n}=\frac{(-1)^n}{\sqrt n}-\frac1n+O\left(\frac {1}{n^{3/2}} \right).$$

Can you take it from here?

• I see where you're going with this, but I'm not totally clear on what you did in the second equality above. How was that $(-1)^n$ cancelled out on the second term of the expansion?
– Noa
Dec 16, 2017 at 11:17
• @OlivierOloa, why so comlicated? Why not simply set: $\sin(x)=O(1)$ and $\frac1{1+x}=1+O(x)$. Then, as $n\to\infty$, we immediately see,$$(-1)^n\frac {\sqrt n}{(-1)^n+\sqrt n}\:\sin \frac {1}{\sqrt n}=(-1)^n\frac {1}{1+\frac{(-1)^n}{\sqrt n}}\:\sin \frac {1}{\sqrt n}=(-1)^n+O\left( \frac{1}{\sqrt n}\right).$$ Clearly, $\sum\limits_{n=2}^{\infty}(-1)^n+O\left( \frac{1}{\sqrt n}\right)$ can't be convergent. Nov 16, 2020 at 20:16
• @OlivierOloa,$$(-1)^n\frac{1}{1+\frac{(-1)^n}{\sqrt{n}}}\sin\frac{1}{\sqrt{n}}=(-1)^n\cdot\left(1+O\left(\frac{1}{\sqrt{n}}\right)\right)\cdot O(1)\\=(-1)^n\cdot\left(1\cdot O(1)+O\left(\frac{1}{\sqrt{n}}\right)\cdot O(1)\right)=\left((-1)^n+(-1)^nO\left(\frac{1}{\sqrt{n}}\right)\right)=(-1)^n+O\left(\frac{1}{\sqrt{n}}\right)$$ Nov 17, 2020 at 11:45
• @Philipp Please observe that with your reasoning one may write $$(-1)^n\:\sin \frac {1}{n^2}=(-1)^n \cdot O\left( 1\right)=(-1)^n$$ and one may conclude that $\sum_{n\ge1}(-1)^n\:\sin \frac {1}{n^2}$ being the same nature of $\sum_{n\ge1}(-1)^n$ is divergent... but in fact $$\left| (-1)^n\:\sin \frac {1}{n^2}\right|\le \frac{1}{n^2}$$ and the series $\sum_{n\ge1}(-1)^n\:\sin \frac {1}{n^2}$ is thus (absloutely) convergent. Nov 19, 2020 at 22:59
• @OlivierOloa, your "counter-example" is wrong. The constant $0$-function is also of the same class as $(-1)^n$, or as you said of the same nature, because $\Big|\frac{0}{(-1)^n}\Big|\leq C$, where $C$ can be any positive constant. But clearly $\sum\limits_{n=1}^{\infty}0$ is convergent yet $\sum\limits_{n=1}^{\infty}(-1)^n$ is divergent. So only because two functions are of the same class it doesn't mean that the two corresponding series have the same convergence behavior. So I still don't see a problem with my approach. Nov 19, 2020 at 23:42

From ${1\over1+u}=1-u+{u^2\over1+u}$ we have

$${\sqrt n\over(-1)^n+\sqrt n}={1\over1+{(-1)^n\over\sqrt n}}=1-{(-1)^n\over\sqrt n}+{{1\over n}\over1+{(-1)^n\over\sqrt n}}=1-{(-1)^n\over\sqrt n}+{1\over n+(-1)^n\sqrt n}$$

Now, on multiplying by $(-1)^n\sin\left(1\over\sqrt n\right)$, we see that

$$\sum_{n=2}^\infty(-1)^n\sin\left(1\over\sqrt n\right)$$

converges conditionally while

$$\sum_{n=2}^\infty{(-1)^n\over n+(-1)^n\sqrt n}\sin\left(1\over\sqrt n\right)$$

converges absolutely (since $\sin(1/\sqrt n)/(n+(-1)^n\sqrt n)\approx1/n^{3/2}$ for large $n$), but

$$\sum_{n=2}^\infty{1\over\sqrt n}\sin\left(1\over\sqrt n\right)\approx\sum_{n=2}^\infty{1\over n}$$

diverges. So the overall sum diverges.

• Wouldn't this solution involve rearranging the terms? The proof you're suggesting encapsulates the assumption that the series itself converges conditionally, so rearranging the terms in that case is meaningless...
– Noa
Dec 19, 2017 at 17:28
• @Noa, it's good you think to ask about this. There actually isn't any rearrangement of terms here (nor is there in Rigel's answer). You can see this by replacing $\infty$ with $N$ as the upper limit of the sum(s). That is, each partial sum splits into three pieces. For two of the pieces the limit of the partial sum exists, while for the third piece it doesn't. That's enough to show the original infinite sum does not converge. Dec 19, 2017 at 19:33
• Thank you for the clarification. This question's really been bugging me for a while.
– Noa
Dec 20, 2017 at 8:25