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.
