Finding a bound for $\sum_{n=k}^l \frac{z^n}{n}$ For $z\in\mathbb{C}$ such that $|z|=1$ but $z\neq1$ and $0<k<l$, I'm trying to prove that:
$$\left|\sum_{n=k}^l \frac{z^n}{n}\right| \leq \frac{4}{k|1-z|}$$
It's more of a game that slowly frustrates  me... I've got 
$$\left|\sum_{n=k}^l \frac{z^n}{n}\right|=\left|\frac{1}{l}\frac{1-z^{l+1}}{1-z}-\frac{1}{k}\frac{1-z^k}{1-z}-\sum_{n=k}^{l-1}\left(\frac{1-z^{n+1}}{1-z}\right)\left(\frac{1}{n+1}-\frac{1}{n}\right)\right|$$
Then I've tried to use the Triangle Inequality over and over again, but I never actually got to the point... Do you have any idea from this point?

Start from 
$$|1-z|\cdot\left|\sum_{n=k}^l \frac{z^n}{n}\right|=\left|\sum_{n=k}^l\frac{z^n}n-\sum_{n=k+1}^{l+1}\frac{z^n}{n-1}\right|.$$
The term for $n=k$ and $n=l+1$ can be bounded by $1/k$. To conclude, notice that 
$$\left|\sum_{n=k+1}^lz^n\left(\frac 1n-\frac 1{n-1}\right)\right|\leqslant \sum_{n=k+1}^l\left|z^n\left(\frac 1n-\frac 1{n-1}\right)\right|=\sum_{n=k+1}^l\left(\frac 1{n-1}-\frac 1n\right)=\frac 1{k}-\frac 1l\leqslant \frac 1k.$$
