# Sum inequality: $\sum_{k=1}^n \frac{\sin k}{k} \le \pi-1$

I’m interested in finding an elementary proof for the following sum inequality:
$$\sum_{k=1}^n \frac{\sin k}{k} \le \pi-1$$

If this inequality is easy to prove, then one may easily prove that the sum is bounded.

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An approach of Abel Summation:

Let

$$S_n=\sum_{i=1}^{n}\sin k,\quad S_0=0.$$

Then

$$\sum_{k=1}^{n}\frac{\sin k}{k}=\sum_{k=1}^{n}\frac{S_k-S_{k-1}}{k}=\frac{S_n}{n}+\sum_{k=1}^{n-1}\frac{S_k}{k(k+1)}.$$

We have

$$S_n = \sum_{k=1}^{n}\sin k = \mathrm{Im}\left(\sum_{k=1}^{n}e^{ik}\right) = \mathrm{Im}\left(e^{i}\frac{1-e^{in}}{1-e^i}\right).$$

Hence
$$|S_n| \leq \left|\frac{1-e^{in}}{1-e^i}\right| \leq \left|\frac{2}{1-e^i}\right| \approx 2.09.$$

Then

$$\left|\sum_{k=1}^{n}\frac{\sin k}{k}\right| = \left|\frac{S_n}{n} + \sum_{k=1}^{n-1}\frac{S_k}{k(k+1)}\right| \leq 2.09 \left(\frac{1}{n} + \sum_{k=1}^{n-1}\frac{1}{k(k+1)}\right) = 2.09,$$

a sharper bound than $\pi-1$.

Let’s first observe that $\sum_{k=1}^\infty u^k/k=-\ln(1-u)$.

If we’re concerned about the convergence radius, we can always replace $u$ with $ue^{-\epsilon}$ and let $\epsilon\rightarrow0$. The branch of $\ln$ we’re using is the one defined on $\mathbb{C}\setminus(-\infty,0]$: i.e. $\ln(re^{i\theta})=\ln r+i\theta$ where $r>0$ and $\theta\in(-\pi,\pi)$.

Inserting $\sin x=(e^{ix}-e^{-ix})/2i$, we get
$$\sum_{k=1}^\infty \frac{\sin kx}{k} =\sum_{k=1}^\infty \frac{e^{ikx}-e^{-ikx}}{2ki} =\frac{\ln(1-e^{-ix})-\ln(1-e^{ix})}{2i}$$
At this point, I have two alternative solutions. In either case, I assume $x\in[0,\pi)$ to help stay within the selected branch of the logarithm.

You can look at the triangle with corners $O=0$, $I=1$ and $A=1-e^{-ix}$: this has $IO=IA$ and $\angle OIA=x$, so $\angle AOI=\frac{\pi-x}{2}$. This makes the imaginary part of $\ln(1-e^{-ix})=\angle AOI=\frac{\pi-x}{2}$; for $\ln(1-e^{ix})$ it is $-\frac{\pi-x}{2}$. The real part of the logarithm cancels out, and what remains is $\frac{\pi-x}{2}$.

Alternatively, while ensuring we stay within the branch of the logarithm, we get
$$\sum_{k=1}^\infty \frac{\sin kx}{k} =\frac{1}{2i}\ln\frac{1-e^{-ix}}{1-e^{ix}} =\frac{\ln(-e^{-ix})}{2i} =\frac{\ln(e^{i(\pi-x)})}{2i} =\frac{\pi-x}{2}.$$

Thus, not only is the sum less than $\pi-1$. It is exactly $\frac{\pi-1}{2}$. And the more general sum
$$\sum_{k=1}^\infty \frac{\sin kx}{k} =\frac{\pi-x}{2}$$
for $x\in[0,\pi]$: if $x=\pi$, the sum becomes zero (either by limit or because all the terms are zero).