Prove $\sum_{n≥1}\frac{\sin n}{n}= \frac{1}{2}(\pi-1)$

Question

I recently learnt that sums of sine and cosine series whose arguments are in arithmetic progression can be evaluated using complex numbers. I modified it a little; like below I divided it by the argument of sine, and didn't expected a closed form sum. But surprisingly , Wolfram calculator gives this nice answer for it's sum. $$\sum_{n≥1}\frac{\sin n}{n}= \frac{1}{2}(\pi-1)$$

I've never encountered such series so I'm interested in knowing that what are the methods using which we can prove these sums ( assuming the result given by calculator as true ). Thanks !

See https://math.stackexchange.com/q/227751/42969 and the linked questions. — Martin R, Sep 26 '23 at 11:57
Also related: https://math.stackexchange.com/q/3361618/42969 — Martin R, Sep 26 '23 at 11:58
@MartinR I tried scrolling but these questions didn't appeared on related questions while I was writing. Thanks! — An_Elephant, Sep 26 '23 at 12:00
@AjinShajiJose I'm a high school dropper .I have heard about that it's used in quantum mechanics but as said , I've only "heard" and not familiar with either of them. I hope to learn in future. Thanks for telling me the technique. — An_Elephant, Sep 26 '23 at 12:02
I found this: https://math.stackexchange.com/questions/13490/proving-that-the-sequence-f-nx-sum-limits-k-1n-frac-sinkxk-is — Bob Dobbs, Sep 26 '23 at 13:56
Is the domain for n assumed to be integers, natural numbers? In other words what is the significance of $n \geq 1$ instead of $\sum_{n=1}^\infty$ — nickalh, Sep 26 '23 at 15:45
And for the record this is a duplicate question right? @MartinR 's linked questions. — nickalh, Sep 26 '23 at 15:47

score 2 · Accepted Answer · answered Sep 26 '23 at 15:35

2

We have that $$\sum_{n=1}^\infty\frac{z^n}n=-\ln(1-z)$$ for $|z|<1.$

I think due to Dirichlet test, we can use this formula for $z=e^{ix}\neq1$. Then,

$$\begin{align} \sum_{n=1}^\infty\frac{\sin nx}n &= \frac1{2i}\left(\sum_{n=1}^\infty\frac{e^{ixn}}n-\sum_{n=1}^\infty\frac{e^{-ixn}}n\right)\\\\ &=\frac1{2i}\ln\left(\frac{1-e^{-ix}}{1-e^{ix}}\right)\\\\ &=\frac{\pi-x}{2} \end{align}$$ which is true for $0<x<2\pi$.

answered Sep 26 '23 at 15:35

Bob Dobbs

10,988

1

You pulled that out of a hat! You have to be explicit with your limits when complex logarithms are concerned. In particular, $|e^{ix}|$ and $|e^{-ix}|$ are not less than $1$, so you can't just plug them into your first equation. – TonyK Sep 26 '23 at 15:50
$\ln1=0$, $\ln(-1)=\pi i$. – Bob Dobbs Sep 26 '23 at 16:41

Prove $\sum_{n≥1}\frac{\sin n}{n}= \frac{1}{2}(\pi-1)$

1 Answers1