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


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 !

  • $\begingroup$ What exactly do you want to study in it? $\endgroup$ Apr 17, 2017 at 23:29
  • $\begingroup$ if it's converge or not $\endgroup$ Apr 17, 2017 at 23:32
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    $\begingroup$ Are you asking whether $\sum_{n=1}^\infty\sin(\cos(n))/n$ converges? $\endgroup$
    – stewbasic
    Apr 17, 2017 at 23:33
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    $\begingroup$ This is an interesting question and honestly I do not understand the downvotes or the closing votes. $\endgroup$ Apr 18, 2017 at 3:11
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    $\begingroup$ @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? $\endgroup$ Apr 18, 2017 at 14:01

2 Answers 2


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.

  • $\begingroup$ 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 $\endgroup$ Apr 18, 2017 at 10:28

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.

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    $\begingroup$ you will loose the reader with a Bessel function, simply say $a_m$ decreases fast enough because $\sin \cos x$ is smooth ? $\endgroup$
    – reuns
    Apr 18, 2017 at 3:19
  • $\begingroup$ @user1952009: of course, that and the irrationality measure of $\pi$ are the key ingredients. $\endgroup$ Apr 18, 2017 at 4:43
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    $\begingroup$ Thank you for reopening this question and providing a great answer! $\endgroup$ Apr 18, 2017 at 14:44

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