I was studying and stumbled across a question that I got right but on shaky basis.

The question was :

Let $a_n = (-1)^nsin(\frac{1}{n^2})$ with various answers possible. With the alternating series test, it was easy to say that $\sum a_n$ converge but there was two possible answers, one that it was only convergent and another that it was absolutely convergent.

I kind of guess it's absolutely convergent since the ratio test would be "infinitely slightly" below one but I'm not sure about how to do it mathematically or wether I got it right purely by luck...

Thansk in advance for your help

  • 2
    $\begingroup$ Hint: $|\sin x|<|x|$. $\endgroup$
    – user1010241
    Jan 10, 2022 at 11:16
  • $\begingroup$ oh well for some reason i didn't thought of it thanks $\endgroup$ Jan 10, 2022 at 11:59

2 Answers 2


It is absolutely convergent as well.
Consider this,
$|a_n| = |sin(\frac{1}{n^2})| \leq \frac{1}{n^2}$
I use the property: $sin(\theta) \leq \theta$
We know that $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges.
So, from comparison test, given series will also converge absolutely.



Doubling the number of terms each time,

$$1+\frac14+\frac19+\frac1{16}+\frac1{25}+\frac1{36}+\frac1{49}+\cdots\frac1{n^2}+\cdots\\< 1+\frac14+\frac14+\frac1{16}+\frac1{16}+\frac1{16}+\frac1{16}+\cdots\underbrace{\frac1{2^{2k}}+\frac1{2^{2k}}+\cdots+\frac1{2^{2k}}}_{2^k}+\cdots\\<1+\frac24+\frac4{16}+\cdots\frac1{2^k}+\cdots<2.$$

Note that the same argument works for convergence of $\dfrac1{n^{1+\epsilon}}$ with $\epsilon>0$. And with a small modification, divergence of $\dfrac1{n}$.

  • $\begingroup$ I don't really understand how I could use that, because yeah I can get a bound for $\frac{1}{n^2}$ but it will not tell me that $sin(\frac{1}{n^2})$ converge since sin(2) is just a constant. $\endgroup$ Jan 10, 2022 at 12:56
  • $\begingroup$ @user15757055: this is a bound for the absolute series ! $\endgroup$
    – user1010241
    Jan 10, 2022 at 14:13
  • $\begingroup$ Oh yeah sorry I don't know why it didn't clicked but it's definitely a proof that the series is absolutely convergent when you have a bound for it ahah sorry again and thanks for your help $\endgroup$ Jan 10, 2022 at 14:44

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