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What would the sum $$\sum_{n=1}^{\infty} \frac{|\sin (n x)|}{n^2}$$ evaluate to?

Without the modulus, Wolfram says that $$\sum_{n=1}^{\infty} \frac{\sin (n x)}{n^2}=\frac{1}{2} i\left(\mathrm{Li}_2\left(e^{-i x}\right)-\mathrm{Li}_2\left(e^{i x}\right)\right)$$.

I have two questions:

1.) How did we get $\sum_{n=1}^{\infty} \frac{\sin (n x)}{n^2}=\frac{1}{2} i\left(\mathrm{Li}_2\left(e^{-i x}\right)-\mathrm{Li}_2\left(e^{i x}\right)\right)$?

2.) What is $\sum_{n=1}^{\infty} \frac{|\sin (n x)|}{n^2}$?

Thank you!

Rócherz
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Anomaly
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    we can conclude that the given series $\sum_{n=1}^{\infty} \frac{|\sin(nx)|}{n^2}$ is convergent, and its sum is bounded above by $\frac{2\pi^2}{6}$, which simplifies to $\frac{\pi^2}{3}$. – user253963 May 10 '23 at 13:57
  • How did you get that @user253963? – Anomaly May 10 '23 at 14:00
  • Since the terms involve absolute values, we know that $|\sin(nx)| \leq 1$ for any value of $x$. – user253963 May 10 '23 at 14:01
  • Why the $2$ @user253963 ? – Bruno B May 10 '23 at 14:02
  • @user253963, can you please give a detailed solution as an answer? What you are saying makes sense. Thank you! – Anomaly May 10 '23 at 14:02
  • The sum $\sum_{n = 1}^\infty \frac{1}{n^2}$ is well-known to be equal to $\frac{\pi^2}{6}$. – Bruno B May 10 '23 at 14:03
  • Yes. But since $|\sin(nx)| \leq 1$, what about the values of less than 1 but greater than 0? – Anomaly May 10 '23 at 14:05
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    What user253963 did is just a standard comparison test between series with non-negative terms: $0 \leq \frac{|\sin(nx)|}{n^2} \leq \frac{1}{n^2}$ with the RHS being summable, thus the middle term is also summable and the sum $\sum_{n = 1}^\infty \frac{|\sin(nx)|}{n^2} $ is lesser than the sum $\sum_{n = 1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$. – Bruno B May 10 '23 at 14:13
  • You could observe that the n=1 (possibly n <= 4) parts are a very large part of the total sum and use this to improve the sum. |sin x| <= sum <= |sin x| + (pi^2 - 6) / 6 instead of 0 <= sum <= pi^2/6. – gnasher729 Jun 15 '23 at 06:31
  • Guys, I think OP wants to know what the series with the absolute value evaluates to and not verify that the series converges. – Accelerator Jun 15 '23 at 07:08

1 Answers1

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To answer your first question, using the identities $\displaystyle{\sin(nx)=\frac{e^{inx}-e^{-inx}}{2i}}$ and $\displaystyle{\frac{1}{i}=-i}$, $$\begin{align*} \sum_{n=1}^\infty\frac{\sin(nx)}{n^2}&=\frac{1}{2i}\sum_{n=1}^\infty\frac{e^{inx}-e^{-inx}}{n^2}\\ &=\frac{i}{2}\left(\sum_{n=1}^\infty\frac{e^{-inx}}{n^2}-\sum_{n=1}^\infty\frac{e^{inx}}{n^2}\right)\\ &=\boxed{\frac{i}{2}\left(\operatorname{Li}_2(e^{-ix})-\operatorname{Li}_2(e^{ix})\right)} \end{align*}$$ where $\operatorname{Li}_2(x)$ is the Dilogarithm defined as, $$\operatorname{Li}_2(x)=\sum_{n=1}^\infty\frac{x^n}{n^2}$$ for $|x|<1$.

bob
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