# Why does this limit make the sum converge?

I was messing around with sums on Wolfram Alpha and I came across a weird result that I do not understand. I initially typed in $$\sum_{n=1}^\infty\sin(nx)$$ to which the response was that it does not converge (which was expected), however, underneath was a result involving a 'Dirichlet Regularization' that stated $$\lim_{s \rightarrow 0}\sum_{n=1}^\infty n^{-s}\sin(nx) = \frac{1}{2}\cot \frac{x}{2}$$.

I've tried to do some research but I have yet to find anything that helps me understand why this limit actually makes this sum converge. Any explaination or direction on where I can learn more about this would be greatly appreciated!

Edit: A proof for this sum as well would be greatly appreciated

• When $s>0$, you get an alternating series with decreasing terms. That makes it conditionally converge. – user65203 Apr 15 '20 at 15:21
• If you are intrigued and want to see a fantastic primary source on results similar to this, check out GH Hardy’s Divergent Series text. – TM Gallagher Apr 15 '20 at 16:12
• @YvesDaoust: This is not an alternating series (nor do the terms decrease monotonically). However, $n^{-s}$ is monotonically decreasing and $\sum_{k=1}^n\sin(kx)$ is bounded, so Dirichlet's Test appliies. – robjohn Apr 15 '20 at 19:38
• @robjohn: I mean with non-constant signs and decreasing envelope. This is for an intuitive explanation. – user65203 Apr 15 '20 at 19:40

Define the Polylogarithm functions $$\newcommand{\Li}{\operatorname{Li}} \Li_s(x)=\sum_{k=1}^\infty\frac{x^k}{k^s}\tag1$$ For $$s\gt0$$, Summation by Parts and the Mean Value Theorem say \begin{align} (1-x)\Li_s(x) &=\sum_{k=1}^\infty\frac{x^k-x^{k+1}}{k^s}\tag{2a}\\ &=x+\sum_{k=2}^\infty x^k\!\left(\frac1{k^s}-\frac1{(k-1)^s}\right)\tag{2b}\\ &=x-s\sum_{k=2}^\infty\frac{x^k}{(k-\theta_{k,s})^{s+1}}\tag{2c} \end{align} Explanation:
$$\text{(2a)}$$: multiply $$(1)$$ by $$1-x$$
$$\text{(2b)}$$: Summation by Parts
$$\text{(2c)}$$: Mean Value Theorem where $$0\lt\theta_{k,s}\lt1$$
Thus, $$\frac1{(k-\theta_{k,s})^{s+1}}$$ is decreasing from something less than $$1$$ to $$0$$, and, as usual, we have the bound $$\left|\,\sum\limits_{k=1}^ne^{ikx}\,\right|\le\frac1{|\sin(x/2)\,|}$$. Dirichlet's Test then says that $$\left(1-e^{ix}\right)\Li_s\left(e^{ix}\right)=e^{ix}+O\!\left(\frac s{|\sin(x/2)\,|}\right)\tag3$$ This means that for $$e^{ix}\ne1$$, \begin{align} \lim_{s\to0^+}\Li_s\left(e^{ix}\right) &=\frac{e^{ix}}{1-e^{ix}}\tag{4a}\\ &=-\frac{e^{ix/2}}{e^{ix/2}-e^{-ix/2}}\tag{4b}\\ &=-\frac{\cos(x/2)+i\sin(x/2)}{2i\sin(x/2)}\tag{4c}\\ &=-\frac12+\frac i2\cot(x/2)\tag{4d} \end{align} Explanation:
$$\text{(4a)}$$: take the limit of $$(3)$$
$$\text{(4b)}$$: multiply numerator and denominator by $$-e^{-ix/2}$$
$$\text{(4c)}$$: write things in terms of $$\sin$$ and $$\cos$$
$$\text{(4d)}$$: simplify
Taking the imaginary parts of $$\text{(4d)}$$, we get $$\bbox[5px,border:2px solid #C0A000]{\lim_{s\to0^+}\sum_{k=1}^\infty\frac{\sin(kx)}{k^s}=\frac12\cot(x/2)}\tag5$$ The original sum, for $$s=0$$, still does not converge, but taking the limit of this regularization gives a value.