For every $ \alpha \in [\ln(4/e),1]$ there is $x\in [-\pi,\pi]$ such that $|\sum_{n=2}^{\infty}\frac{e^{inx}}{n^2-n}|=\alpha$ Consider 
$$g(x):=\sum_{n=2}^{\infty}\frac{e^{inx}}{n^2-n}$$
(This series converges absolutely and uniformly on $\mathbb{R}$.) How do I prove the following claim
$$
\forall \alpha \in [\ln(4/e),1]\,\exists x\in [-\pi,\pi]:\quad |g(x)|=\alpha\quad ?
$$
I have shown that $|g|\leq 1$ on $\mathbb{R}$, and that $$\sum_{n=2}^{\infty}\frac{x^n}{n^2-n}=(1-x)\ln(1-x)+x$$
for $x\in [-1,1)$. I am not sure, but I think that I should show that there exists $x\in [-\pi,\pi]$ such that
$$
\ln(4/e)=\sum_{n=2}^{\infty}\frac{(-1)^n}{n^2-n}\stackrel{?}\leq |g(x)|\leq 1
$$
 A: Note that, since your series converges uniformly, the function $g$ is continuous on $\Bbb R$.
By telescoping, we have $g(1)=\sum_{n=1}^\infty \frac{1}{n^2-n}=1$.
Also, by \eqref{*}, 
\begin{align}
g(-\pi)&=\big(1-\exp(i\pi)\big)\ln\big(1-\exp(i\pi)\big) + \exp(i\pi)=2\ln(2)-1=\ln(4)-\ln(e)=\ln\left(\frac4e\right).
\end{align}
Thus, by the intermediate value Theorem, $$[g(-\pi),g(1)]=\left[\ln\left(\frac4e\right),1\right]\subseteq g([-\pi,1])\subseteq g([-\pi,\pi]).$$
From this immediately follows your claim (note that I actually proved something stronger than your claim.)

For completeness, I will state a proof for the fact you already mentioned:
\begin{equation}\tag{*}\label *\sum_{n=2}^{\infty}\frac{x^n}{n^2-n}=(1-x)\ln(1-x)+x \quad \text{for } x\in[-1,1[.\end{equation}
Proof sketch: For $x=-1$, you have to do some straight-forward calculations using the Maclaurin expansion of $\ln(1-x)$. For $x\in]-1,1[$, we have (with absolute convergence; and uniform convergence on any compact interval contained in $]-1,1[$): $$\ln(1-x)=-\sum_{n=1}^\infty \frac{x^n}n,$$
and thus 
$$(1-x)\ln(1-x)+x=x+\sum_{n=1}^\infty \frac{x^{n+1}}n-\frac{x^n}n=x-x+\sum_{n=2}^\infty \frac{x^n}{n-1}-\frac{x^n}n,$$
from which follows the claim since $\frac{1}{n-1}-\frac1n=\frac1{n^2-n}$.
