I have a series defined like this: $$\sum_{n=1}^{\infty} (-1)^n \left(\cos \frac{1}{n}\right)^n$$

and I need to find out whether it converges or diverges. I know that $\lim_{n\to\infty} |a_n| = 1$ but does it tell something about the series when the sign is alternating?

  • 3
    $\begingroup$ $a_n\to 0 \iff |a_n|\to 0$. $\endgroup$
    – Galc127
    May 30, 2016 at 12:43

2 Answers 2


An infinite series $\sum\limits_{n} a_n$ will never converge if $a_n \not\to 0$. It doesn't matter if it is alternating or not.

  • $\begingroup$ Thanks, I didn't know this holds. $\endgroup$
    – Gogis
    May 30, 2016 at 12:45
  • $\begingroup$ You're welcome. See for example on Wikipedia Term test. $\endgroup$
    – Eff
    May 30, 2016 at 12:47
  • $\begingroup$ @Gogis It's true, and fairly easy to show (it also makes pretty good exercise) $\endgroup$
    – 5xum
    May 30, 2016 at 12:50

The original series diverges, as you end-up subtracting then adding $1$.

Anyway, by grouping the terms in pairs, you get a new series with the general term

$$\left(\cos\left(\frac1{2n}\right)\right)^{2n}-\left(\cos\left(\frac1{2n+1}\right)\right)^{2n+1}\\ =\left(1-\frac1{2(2n)^2}+\cdots\right)^{2n}-\left(1-\frac1{2(2n+1)^2}+\cdots\right)^{2n+1}\\ =1-\frac{2n}{8n^2}+\cdots-1+\frac{2n+1}{2(2n+1)^2}-\cdots\\ \approx-\frac1{2n(2n+1)}=\Theta\left(\frac1{n^2}\right),$$

which does converge, I guess.


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