<s>[INST] Evaluating $\sum_{k=1}^{n} \frac{1}{k(k+1)}$ I have just started learning sums. I need to evaluate the following sum:
$$S_{n} = \sum_{k=1}^{n} \frac{1}{k(k+1)}$$
$$a_{1} = \frac{1}{2}, a_{2} = \frac{1}{6}, a_{3} = \frac{1}{12}, a_{4} = \frac{1}{20}, a_{5} = \frac{1}{30}, a_{6} = \frac{1}{42}, ...$$
I tried splitting it to partial fractions:
$$\frac{1}{k(k+1)} = \frac{A}{k} + \frac{B}{k+1}$$
$$1 = A(k+1) + Bk$$
$$1 = Ak + A + Bk$$
$$\begin{cases} A = 1 \\ 0 = A + B \quad \Rightarrow B = -1\end{cases}$$
$$\Longrightarrow S_{n} = \sum_{k=1}^{n} \Big( \frac{1}{k} - \frac{1}{k+1} \Big)$$
$$S_{n} = \sum_{k=1}^{n} \frac{1}{k} - \sum_{k=1}^{n} \frac{1}{k+1}$$
Wolframalpha gives answer $S_{n} = \frac{n}{n+1}$
Help's appreciated. P.S: Are the tags okay?
 [/INST] $$S_{n} = \sum_{k=1}^{n} \frac{1}{k}-\frac{1}{k+1}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{n}+\frac{1}{n}-\frac{1}{n+1}=1-\frac{1}{n+1}$$
Check this: https://en.wikipedia.org/wiki/Telescoping_series
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