## 21 March, 2015

### Does the Trigonometric Harmonic Series Converge?

It is well known that the harmonic series $H(x)=\sum_{n=1}^{\infty} xn^{-1}=+\infty$ for every $x\neq0$, but what about the trigonometric harmonic series $T(x)=\sum_{n=1}^{\infty}e^{inx}n^{-1}$?  Obviously for $k=1,2,\ldots$ we have $T(2k\pi)=H(1)=+\infty$.  It is an interesting fact that the cancellation properties inherent in $T$ imply convergence.  This is relatively straight-forward to prove this using a modificaton of Leibniz's alternating series test.  More remarkable is that the convergence is actually absolute.

In order to investigate the convergence of
$$(1)\;\;\;\;\;T(x)=\sum_{n=1}^{\infty}\frac{e^{inx}}{n}<\infty,$$
first note that
$$\lim_{n\to\infty}|z^{n}|\to0$$
for every $z\in\mathbb{C}$ with $|z|<1$.  Since
$$1>\frac{1}{n}>\frac{1}{n+1}>0$$ for all $n>1$, we find $\frac{1}{n}\searrow0$ (monotonically decreases to zero) and so Dirichlet's test implies
$$(2)\;\;\;\;\;\sum\limits_{n=1}^{\infty}\frac{z^{n}}{n}<\infty,$$
the convergence taking place and being absolute for every $z$ with $|z|<1$.  To deal with the boundary $|z|=1$, note that if $|z|=1$ and $z\neq1$ (i.e. $z\neq1+0i$), then we have
$$\left|\sum_{n=1}^{N}z^{n}\right|=\left|\frac{1-z^{N+1}}{1-z}\right|\leq\frac{2}{1-z}<\infty.$$
The upper bound $M=\frac{2}{1-z}$ is independent of $N$ and so (2) holds for all $|z|\leq1$, except when $z=1$.  Putting $z\mapsto e^{inx}$ shows that (1) converges absolutely for every $x\neq 2k\pi$ ($k=1, 2, \ldots$).

To carry out the actual summation for $T(x)$ is a tedious exercise in complex analytic methods, and the resulting formulas are unworkable (although again rather remarkably, they contain only elementary functions).  Another approach is to recognize that $T(x)$ is the Fourier transform (series) of some periodic function with Fourier coefficients $\hat{f}(0)=0$ and for $n>1$
$$\hat{f}(n)=\frac{1}{n}.$$
Despite this, the computation is relatively straight-forward for certain values of $x$.  For example, take $x=1$ and note that
$$T(1)=\sum_{n=1}^{\infty}\frac{e^{in}}{n}.$$
Writing
$$\int\left(\underbrace{(e^{iz})^{1}+(e^{iz})^{2}+\ldots}_{\text{geomtric series with ratio }r=e^{iz}}\right)dz=\int\frac{e^{iz}}{1-e^{iz}}\;dz,$$
we find that (with $u=1-e^{iz}$)
$$\sum_{n=1}^{\infty}\frac{1}{in}(e^{iz})^{n}=i\int\frac{du}{u}=i\ln(1-e^{iz}).$$
Combining all of this together, we obtain
\begin{align*} T(1) &=\left(i\ln(1-e^{iz})\right)\Big|_{z=1}\\ &=i\ln\left(e^{i/2}\left((e^{-i/2}-e^{i/2}\right)\right)\\ &=i\ln\left(e^{i/2}\right)+i\ln\left(2i\sin\left(-\frac{1}{2}\right)\right)\\ &=-\frac{1}{2}+i\left(\ln(-i)+\ln\left(\sin\left(\frac{1}{2}\right)\right)\right)\\ &=-\frac{1}{2}+\frac{\pi}{2}+i\ln\left(2\sin\left(\frac{1}{2}\right)\right) \end{align*}
Since
$$T(1)=\sum_{n=1}^{\infty}\left(\frac{\cos n}{n}+i\frac{\sin n}{n}\right),$$
taking real and imaginary parts yields
$$T(1)=\frac{-\ln(2-2\cos(1))}{2}+i\frac{\pi-1}{2}.$$

The graphic at the beginning of the post shows the graph of $\sin n/n$ on the $(n,x)$ plane.