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Why does this series $$\sum\limits_{n=2}^\infty\frac{\cos(n\pi/3)}{n}$$ converge? Can't you use a limit comparison with $1/n$?

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  • $\begingroup$ Can you please edit the title so it is clear what you are asking? Is it $\sum_{n=1}^{\infty} \cos \dfrac{n\pi}{3n}$ $\endgroup$ – Aryabhata Apr 5 '13 at 6:04
  • $\begingroup$ Careful, your first statement is not correct. The cosine term will oscillate as $n$ gets large and never approach a single value. $\endgroup$ – Jared Apr 5 '13 at 6:04
  • $\begingroup$ @Jared you are right, will correct it. Is it an alternating series? $\endgroup$ – Billy Thompson Apr 5 '13 at 6:07
  • $\begingroup$ Yes, this sounds like the way to go. It's not a strictly alternating series, but the sign change in cosine is what causes convergence. You may even be able to evaluate this explicitly using telescoping series, because we know the values of $\cos(\frac{n\pi}{3})$ for all integral $n$. $\endgroup$ – Jared Apr 5 '13 at 6:08
  • $\begingroup$ @Jared is there any other way to evaluate it? $\endgroup$ – Billy Thompson Apr 5 '13 at 6:10
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First of all your conclusion is wrong since $\lim_{n \to \infty} \cos(n \pi/3)$ doesn't exist.

The convergence of $$\sum_{n=1}^N \dfrac{\cos(n\pi/3)}{n}$$ can be concluded based on Abel partial summation (The result is termed as generalized alternating test or Dirichlet test). We will prove the generalized statement first.

Consider the sum $S_N = \displaystyle \sum_{n=1}^N a(n)b(n)$. Let $A(n) = \displaystyle \sum_{n=1}^N a(n)$. If $b(n) \downarrow 0$ and $A(n)$ is bounded, then the series $\displaystyle \sum_{n=1}^{\infty} a(n)b(n)$ converges.

First note that from Abel summation, we have that \begin{align*}\sum_{n=1}^N a(n) b(n) &= \sum_{n=1}^N b(n)(A(n)-A(n-1))\\&= \sum_{n=1}^{N} b(n) A(n) - \sum_{n=1}^N b(n)A(n-1)\\ &= \sum_{n=1}^{N} b(n) A(n) - \sum_{n=0}^{N-1} b(n+1)A(n) \\&= b(N) A(N) - b(1)A(0) + \sum_{n=1}^{N-1} A(n) (b(n)-b(n+1))\end{align*} Now if $A(n)$ is bounded i.e. $\vert A(n) \vert \leq M$ and $b(n)$ is decreasing, then we have that $$\sum_{n=1}^{N-1} \left \vert A(n) \right \vert (b(n)-b(n+1)) \leq \sum_{n=1}^{N-1} M (b(n)-b(n+1))\\ = M (b(1) - b(N)) \leq Mb(1)$$ Hence, we have that $\displaystyle \sum_{n=1}^{N-1} \left \vert A(n) \right \vert (b(n)-b(n+1))$ converges and hence $$\displaystyle \sum_{n=1}^{N-1} A(n) (b(n)-b(n+1))$$ converges absolutely. Now since $$\sum_{n=1}^N a(n) b(n) = b(N) A(N) + \sum_{n=1}^{N-1} A(n) (b(n)-b(n+1))$$ we have that $\displaystyle \sum_{n=1}^N a(n)b(n)$ converges.

In your case, $a(n) = \cos(n \pi/3)$. Hence, $$A(N) = \displaystyle \sum_{n=1}^N a(n) = - \dfrac12 - \cos\left(\dfrac{\pi}3(N+2)\right)$$which is clearly bounded.

Also, $b(n) = \dfrac1{n}$ is a monotone decreasing sequence converging to $0$.

Hence, we have that $$\sum_{n=1}^N \dfrac{\cos(n\pi/3)}{n}$$ converges.


Look at some of my earlier answers for similar questions.

For what real numbers $a$ does the series $\sum \frac{\sin(ka)}{\log(k)}$ converge or diverge?

Give a demonstration that $\sum\limits_{n=1}^\infty\frac{\sin(n)}{n}$ converges.

If the partial sums of a $a_n$ are bounded, then $\sum{}_{n=1}^\infty a_n e^{-nt}$ converges for all $t > 0$


If you are interested in evaluating the series, here is a way out. We have for $\vert z \vert \leq 1$ and $z \neq 1$, $$\sum_{n=1}^{\infty} \dfrac{z^n}n = - \log(1-z)$$ Setting $z = e^{i \pi/3}$, we get that $$\sum_{n=1}^{\infty} \dfrac{e^{in \pi/3}}n = - \log(1-e^{i \pi/3})$$ Hence, \begin{align} \sum_{n=1}^{\infty} \dfrac{\cos(n \pi/3)}n & = \text{Real part of}\left(\sum_{n=1}^{\infty} \dfrac{e^{in \pi/3}}n \right)\\ & = \text{Real part of} \left(- \log(1-e^{i \pi/3}) \right)\\ & = - \log(\vert 1-e^{i \pi/3} \vert) = 0 \end{align} Hence, $$\sum_{n=2}^{\infty} \dfrac{\cos(n \pi/3)}n = - \dfrac{\cos(\pi/3)}1 = - \dfrac12$$

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    $\begingroup$ I think it is time to write a general answer for generalized alternating test and close tons of similar questions as abstract duplicates. $\endgroup$ – user17762 Apr 5 '13 at 6:13
  • $\begingroup$ I support the motion. $\endgroup$ – Did Apr 5 '13 at 7:10
  • $\begingroup$ I'm a little confused about something. On the one hand, you show that $\sum a_n b_n$ should converge absolutely. But $\sum \cos (n \pi 3) n^{-1}$ does not converge absolutely, as the nth term is at least $\frac{1}{2n}$ in absolute value. So I feel I must be missing something? $\endgroup$ – davidlowryduda Apr 5 '13 at 8:14
  • $\begingroup$ @mixedmath Yes, you are absolutely right. $\sum a_n b_n$ doesn't converge absolutely. I have changed it. The proof remains un affected. What we have is $\sum A(n)(b(n) - b(n+1))$ converges absolutely and this is what we want. This doesn't mean that $\sum a(n)b(n)$ converges absolutely. Thanks for pointing this out. $\endgroup$ – user17762 Apr 5 '13 at 15:16
  • $\begingroup$ Awesome. Thanks, and great writeup! $\endgroup$ – davidlowryduda Apr 5 '13 at 16:03
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Note that $$\cos(n\pi/3) = 1/2, \ -1/2, \ -1, \ -1/2, \ 1/2, \ 1, \ 1/2, \ -1/2, \ -1, \ \cdots $$ so your series is just 3 alternating (and convergent) series inter-weaved. Exercise: Prove that if $\sum a_n, \sum b_n$ are both convergent, then the sequence $$a_1, a_1+b_1, a_1+b_1+a_2, a_1+b_1+a_2+b_2, \cdots $$ is convergent. Applying that twice proves your series converges.

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