Evaluate the sum $$\sum_{k=1}^n{2k+1\over k(k+1)}$$ in closed form. You are allowed to have the harmonic number $$H_n = \sum_1^nk^{-1}$$ in your closed–form formula.
I am having trouble with this problem. As of now, I have tried the perturbation method of splitting the last term on one side and the first term on the other. However, that method does not seem to work and I am stuck. It would be much appreciated if you can help me work through this problem.
Using @Galc127's hint:
$\begin{align} \sum_{1 \le k \le n} \frac{2 k + 1}{k (k + 1)} &= \sum_{1 \le k \le n} \left(\frac{1}{k} + \frac{1}{k + 1}\right) \\ &= \sum_{1 \le k \le n} \frac{1}{k} + \sum_{1 \le k \le n + 1} \frac{1}{k} - 1 \\ &= 2 \sum_{1 \le k \le n} \frac{1}{k} + \frac{1}{n + 1} - 1 \\ &= 2 H_n + \frac{1}{n + 1} - 1 \end{align}$
In general, if you have a rational function to sum it is often useful to split into partial fractions.