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

Question

Question_

Does $$\sum_{n=1}^{\infty} n \sin\left({1\over n}\right)$$ converge or diverge?

When I use the Taylor series: $$\sin\left({1 \over x}\right)=\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{(2k-1)!x^k}$$ $$x\sin\left({1 \over x}\right)=1-{1\over 3!x^2}+{1\over5!x^4}-\cdots$$ Although $$\sum_{n=1}^{\infty}{1\over x^m}$$ converges if $m>1$, $\sum{1}$ diverges, so it is impossible to explicit all the terms. Then, how can I check the convergence(or divergence) of the series? Also, I've got stuck in a similar series: $$\sum_{n=1}^{\infty} n \tan\left({1\over n}\right)$$ Could you please help me to get in the right way? Thanks.

score 7 · Accepted Answer · answered Feb 15 '20 at 07:14

7

Since$$\lim_{n\to\infty}n\sin\left(\frac1n\right)=\lim_{n\to\infty}\frac{\sin\left(\frac1n\right)}{\frac1n}=1,$$your series diverges.

answered Feb 15 '20 at 07:14

José Carlos Santos

427,504

That's also simple and nice. Thanks:D – ToBY Feb 15 '20 at 07:16
What test of convergence did you use? – João Víctor Melo Nov 07 '21 at 14:01
@JoãoVíctorMelo The comparison test. – José Carlos Santos Nov 07 '21 at 14:28
I still don't understand well how you used it. – João Víctor Melo Nov 07 '21 at 14:34
According to that test, if you have two series $\sum_{n=1}^\infty a_n$ and $\sum_{n=1}^\infty b_n$ of numbers greater than $0$ and if $\lim_{n\to\infty}\frac{a_n}{b_n}\in(0,\infty)$, then either both series converges or they both diverge. In this case, since the limit is $1$ and since the series $\sum_{n=1}^\infty\frac1n$ diverges, the series $\sum_{n=1}^\infty\sin\left(\frac1n\right)$ diverges too. – José Carlos Santos Nov 07 '21 at 14:54
I've just seen now after I got the answer. – João Víctor Melo Nov 07 '21 at 15:42
@JoséCarlosSantos Actually, the series in the OP's post is $\sum n\sin\left(\frac{1}{n}\right)$ : so your answer is perfectly valid, but you don't use the comparison test, only the fact that the general term of a convergent series must tend to $0$. – TheSilverDoe Nov 07 '21 at 19:02
1

@TheSilverDoe You're right! I didn't even bother to re-read the question. – José Carlos Santos Nov 07 '21 at 19:05

Clement Yung · Answer 2 · 2020-02-15T07:14:28.373

2

Since $\sin{x} \geq \frac{x}{2}$ for $0 \leq x < \frac{\pi}{2}$: $$ \sum_{n=1}^\infty n\sin\left(\frac{1}{n}\right) \geq \sum_{n=1}^\infty n\left(\frac{1}{2n}\right) = \sum_{n=1}^\infty \frac{1}{2} = \infty $$ For $\sum_{n=1}^\infty n\tan\left(\frac{1}{n}\right)$, we have $\tan{x} \geq x$ for $0 \leq x \leq \frac{\pi}{2}$.

edited Feb 15 '20 at 07:14

answered Feb 15 '20 at 07:12

Clement Yung

8,347

That's simple and nice. Thanks:D – ToBY Feb 15 '20 at 07:14

score -2 · Answer 3 · answered Nov 07 '21 at 15:41

The comparisson test says:

if $\lim_{n\to\infty} \frac{a_n}{b_n} = c$, and c is finite and positive, both $\sum_{n = 0}^{\infty} a_n$ and $\sum_{n = 0}^{\infty} b_n$ diverge or both converge.

So, in our case, take $a_n = n \sin(1/n)$ and $b_n = n$, we conclude that since $\lim_{n \to\infty} a_n/b_n = \infty$ then we conclude that the $a_n$ series diverges since the series for $b_n$ diverges, and this is given because:

if $\lim_{n \to\infty} a_n/b_n = \infty,$ if $\sum_{n = 0}^{\infty} b_n$ diverges, then $\sum_{n = 0}^{\infty} a_n$ diverges.

And to complete it:

if $\lim_{n \to\infty} a_n/b_n = 0,$ if $\sum_{n = 0}^{\infty} b_n$ converges, then $\sum_{n = 0}^{\infty} a_n$ converges.

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

3 Answers3