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I was doing some exercises and this one just stunned me. I had to study the convergence of this serie:

$$\sum_{n=1}^{\infty}\frac{\sin(\cos(n))}{n}$$

I tried alot of diffrent things, but I got no where. Can anyone please help me with an idea or a clue just to start ?

Thanks in advance !

Sungjin Kim
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Hptunjy Prjkeizg
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  • What exactly do you want to study in it? – The Dead Legend Apr 17 '17 at 23:29
  • if it's converge or not – Hptunjy Prjkeizg Apr 17 '17 at 23:32
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    Are you asking whether $\sum_{n=1}^\infty\sin(\cos(n))/n$ converges? – stewbasic Apr 17 '17 at 23:33
  • yes exactly, that's what im looking for – Hptunjy Prjkeizg Apr 17 '17 at 23:34
  • I believe it converges, which follows from the Taylor expansion of $\sin(x)$. – Simply Beautiful Art Apr 17 '17 at 23:47
  • your answer doesn't realy help, can u make it more detailed ? – Hptunjy Prjkeizg Apr 18 '17 at 00:01
  • @SimplyBeautifulArt What do you mean? Are you implying that because the $\sum_{n=1}^\infty \frac{\cos(n)^{2k+1}}{n}$ converges and the Taylor Series for $\sin(\cos n)$ will include only odd terms of cosine that we can conclude convergence? I would be careful here, because we have a double summation... even though each term of the Taylor Series converges, that does not mean they converge when all added together. If I understand you right though, that idea could work. – Brevan Ellefsen Apr 18 '17 at 01:06
  • @HptunjyPrjkeizg Here in math stack exchange we don't simply do homework for you. You should show the steps you used to solve the question. Then others will respond. – Arbuja Apr 18 '17 at 01:08
  • @BrevanEllefsen Yes, the lack of answers below makes it clear I do not feel safe about this. :-) But I am heading to bed. Best of luck if you want to make it work, but it seems quite difficult. – Simply Beautiful Art Apr 18 '17 at 01:09
  • @SimplyBeautifulArt haha, fair enough. I might take a stab at it tonight or tomorrow if you don't. Sleep well! – Brevan Ellefsen Apr 18 '17 at 01:13
  • To use Dirichlet's test it would suffice to find a bound $B_k$ for $\left|\sum_{n=1}^N\cos(n)^{2k+1}\right|$ such that $\sum_k B_k/(2k+1)!$ is finite. Each of those sums is bounded (independent of $N$), though they might be quite large when $e^{(2k+1)i}$ is close to $1$. – stewbasic Apr 18 '17 at 01:17
  • this is not a home work i can garentee that, i just found this on internet and wanted to know how to solve it that's all, and thx alot for your help guys ! – Hptunjy Prjkeizg Apr 18 '17 at 01:30
  • Numerically, according to Wolfy, it appears to about 0.0648. – marty cohen Apr 18 '17 at 02:00
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    This is an interesting question and honestly I do not understand the downvotes or the closing votes. – Jack D'Aurizio Apr 18 '17 at 03:11
  • @JackD'Aurizio while it was an interesting question, that alone does not determine whether or not a question is of good quality. (And I also feel like some people downvoted it because the question was too hard, but don't quote me on that) – Simply Beautiful Art Apr 18 '17 at 11:00
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    @SimplyBeautifulArt: of course, that is a matter of personal opinion. Mine is that an interesting question and a good formatting should grant for quality, so I am voting for reopening. Do we really want to be flooded by highly upvoted questions like prove that $\sum_{n\geq 1}\frac{1}{n^2 e^n e^{e^n}}$ is convergent and banish this kind of deeper math? – Jack D'Aurizio Apr 18 '17 at 14:01
  • some think that it's my home work, it's not true i just found this in another forum i can post it if people think that im just asking them to do my work for me – Hptunjy Prjkeizg Apr 18 '17 at 14:43
  • @JackD'Aurizio of course I feel the same. Just guessing at what others think since they'll probably never reveal themselves. – Simply Beautiful Art Apr 18 '17 at 14:59
  • @SimplyBeautifulArt I think this is an interesting problem. I knew that this has some connection to the irrationality measure of $\pi$ and this question that I answered. In my answer $O(n^{-1/\mu+\epsilon})$ is used, and it is great to see that a stronger bound $O(n^{-1/(\mu-1)+\epsilon})$ is possible in the answer. – Sungjin Kim Apr 18 '17 at 18:46

2 Answers2

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I think it is more practical to expand $\sin\cos(x)$ as a Fourier cosine series. We have

$$\sin\cos(x) = 2\sum_{m\geq 0}(-1)^m J_{2m+1}(1) \cos((2m+1)x) $$ where the coefficients $J_{2m+1}(1)$, depending on a modified Bessel function of the first kind, have an exponential decay. On the other hand $$ \sum_{n\geq 1}\frac{\cos(nx)}{n}=-\log\left|2\sin\frac{x}{2}\right|$$ for any $x\not\in 2\pi\mathbb{Z}$, and $-\log\left|2\sin\frac{x}{2}\right|$ cannot be too large for some small $x\in\mathbb{N}$ since $\pi$ has a finite irrationality measure. By exploiting the exponential decay of the previous coefficients we may conclude that the original series is convergent.

Jack D'Aurizio
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Here is an alternate to Jack D'Aurizio's answer, which interestingly also involves the irrationality measure of $\pi$ (call it $\mu$). Suppose $f$ is a $2\pi$-periodic smooth function, such that $\int_0^{2\pi}f(x)\;dx=0$. Let $$ s_n=\frac1n\sum_{k=1}^n f(k). $$ By the Koksma–Hlawka inequality and this bound, $$ |s_n|=O(n^{-1/(\mu-1)+\epsilon}) $$ for any $\epsilon>0$. In particular, choosing $\epsilon<\frac1{2(\mu-1)}$, we have $|s_n|=O(n^{-\epsilon})$. Using summation by parts, $$\begin{eqnarray*} \sum_{n=1}^N\frac{f(n)}n &=&s_N+\sum_{n=1}^{N-1}ns_n\left(\frac1n-\frac1{n+1}\right)\\ &=&s_N+\sum_{n=1}^{N-1}\frac{s_n}{n+1}. \end{eqnarray*}$$ The above implies $s_N\to0$ and the last sum is absolutely convergent, so the LHS converges. In particular setting $f(x)=\sin(\cos(x))$, the series in question converges.

stewbasic
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  • Thank you ! i really liked your answer, i actually thought about using the summation by part, but from the begining, that's why i got no where with it XD – Hptunjy Prjkeizg Apr 18 '17 at 10:28