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Let a series be given as $$\sum_{m=2}^{\infty}\frac{\cos(\pi m)}{\ln(\ln(m))}$$

Is it converges conditionally, converges absolutely or diverges.

Attempt

From computation I see that it does indeed converge (and I can show that it converges using alternating series test / Leibnitz). The thing that is left is to show that the $$\sum_{m=2}^{\infty}\left | \frac{\cos(\pi k)}{\log(\log(k))} \right |$$ diverges and thus that it does not converge absolutely but only conditionally. But trying to use that if $a_n$ does not converge to $0$ then the series diverges I get stock as I see that the sequence $a_n$ does converge to $0$.

1 Answers1

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For $m\in\mathbb{Z}$ we have $\cos(\pi m)=(-1)^m$, so we may write $$\sum_{m=2}^{\infty}\left | \frac{\cos(\pi m)}{\log(\log(m))} \right |=\sum_{m=2}^{\infty}\left | \frac{(-1)^m}{\log(\log(m))} \right|=\sum_{m=2}^{\infty}\frac{1}{\lvert\log(\log(m))\rvert}.$$ Also observe that $\log\log m <m$ for $m>1$. So a lower bound is the harmonic series.

vitamin d
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  • So the lower bound being divergent means that this is divergent, right? –  May 20 '21 at 18:00
  • Yes, $\infty<A\Rightarrow A\to\infty$. Lower bound is used loosely here since both series are divergent. – vitamin d May 20 '21 at 18:03
  • Hmm but itsn't it significant that it should indeed be a lower bound or else this is inconclusive? –  May 20 '21 at 18:04
  • @mathstudent23 I fixed the typo. Now the inequality sign is correct. – vitamin d May 20 '21 at 18:07
  • Why does the inequaility for $\log\log(m)<$ for $m>1$ hold? –  May 20 '21 at 18:35
  • Take a look at this post. You can use analogous arguments for $\log\log m$ instead of $\log m$. (Or observe that $\log m<m$ implies $\log\log m< \log m$ and from both inequalities that $\log\log m <m$ ) – vitamin d May 20 '21 at 18:46
  • @mathstudent23 I'm not quite sure what you mean. I have not used "$n$" in my answer. If you mean $m$: Not one of the statements in my answer ($\cos(\pi m)=(-1)^m, \log\log m<m, \log m<m$) do not hold for $m=3$. – vitamin d May 25 '21 at 17:21
  • My confusion is with this. The series $\sum_{m=2}^{\infty}\frac{1}{\log(\log(m))}$ is not positive for all $m$ which we demanded by taking the absolut value? take $m=2$ for example. Is this not a problem? –  May 25 '21 at 17:24
  • @mathstudent23 The series, which you wrote is positive. You wanted to prove that the series does not converge absolutely and the answer above provides a proof. – vitamin d May 25 '21 at 17:25
  • I meant take $m=2$ which gives the approx $-2.7$. –  May 25 '21 at 17:27
  • Note that the series runs over $m$, so $m$ does not have one value. Are you talking about the lower bound? If that's the case, still not true. – vitamin d May 25 '21 at 17:32
  • I am talking about the lower bound yes since it is demanded that $b_n\leq a_n$ for alle $n \in \mathbb{N}$ to compare. The series of $b_n$ converges if the series $a_n$ converges. the series of $a_n$ diverges if the series of $b_n$ diverges –  May 25 '21 at 17:34
  • @mathstudent23 It is not clear what you're asking. Throwing with expressions $a_n,b_n$ and $m=2$ and $n$ is not understandable at all. Write a clear comment, maybe use Google Translate. – vitamin d May 25 '21 at 17:37
  • @mathstudent23 Should I answer your new question too? – vitamin d May 25 '21 at 17:37
  • My confusion is just from the fact that I am unsure on the fact that if my series is from the index $n=2$ can I then consider $n=3$ without losing properties of convergence etc. –  May 25 '21 at 17:39
  • @mathstudent23 This is not a problem at all. Convergence is about the "last" terms of an infinite series and not about the first ones. You can write $$\sum_{n=2}^{\infty}a_n=a_2+\sum_{n=3}^{\infty}a_n$$ to study convergence of a series, which starts at $3$. – vitamin d May 25 '21 at 17:41
  • My book states that: let $\sum_{n=2}^{\infty}a_n$ and $\sum_{n=2}^{\infty} b_n$ be 2 positive series with $b_n\leq a_n$ for all $n \in \mathbb{N}$. Then $\sum_{n=2}^{\infty}b_n$ converges if $\sum_{n=2}^{\infty}a_n$ converges. $\sum_{n=2}^{\infty}a_n$ diverges if $\sum_{n=2}^{\infty}b_n$ diverges. –  May 25 '21 at 17:47
  • Now, the "for all $n$" seems to me to destroy this argument for divergence since comparing with the harmonic series terms $1/n$ for $n=2$ obviously fails as $\frac{1}{\ln(ln(n))}$ is negativ. –  May 25 '21 at 17:50
  • @mathstudent23 Absolutely nothing "destroys" the argument. I thought it was a trivial step so I didn't write it out. – vitamin d May 25 '21 at 17:52
  • so just let $n \geq 3$? –  May 25 '21 at 17:52
  • @mathstudent23 In order to use the comparison test, $n\ge3$ yes (of course) but the harmonic series is still a "lower bound" for the series if $n\ge2$. This step is trivial and even explained in my comment with the two big sums. – vitamin d May 25 '21 at 17:56
  • But the inequality in your post says $\ln(\ln(x))<x$ but I have to consider $|\ln(\ln(x))|<x$ for $x>1$. Graphing thoes I see that i need $x>1.31$. –  May 25 '21 at 18:53
  • @mathstudent23 I understand what you mean and you are correct. Nonetheless the inequality in my post is still true but is only applied for $m\ge3$ (we can omit the absolut value) since I skipped the trivial step of neglecting the constant at $m=2$. – vitamin d May 25 '21 at 19:21
  • But still. Consider using alternating test for the series in question to prove convergence. You can not do that with $n=2$. That is why I am wondering how that would look if I choose $n=3$. –  May 26 '21 at 16:38
  • Let us continue this discussion in chat. – vitamin d May 26 '21 at 16:52