Evaluating $\sum_{k=1}^{n} \frac{1}{k(k+1)}$

Question

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?

Answer 1

We have \begin{split}\sum_{k=1}^{n} \frac{1}{k(k+1)} = \sum_{k=1}^{n} \frac1k - \frac1{k+1}&=\left(1-\frac12\right)+\left(\frac12-\frac13\right)+\dots+\left(\frac1n-\frac1{n+1}\right)\\&=1-\frac1{n+1}=\frac n{n+1}.\end{split}

Answer 2

$\require{cancel}$Note that$$S_n=1-\cancel{\frac12}+\cancel{\frac12}-\cdots-\cancel{\frac1n}+\cancel{\frac1n}-\frac1{n+1}=1-\frac1{n+1}.$$

Answer 3

$$S_{n} = \sum_{k=1}^{n} \frac{1}{k}-\frac{1}{k+1}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{n}+\frac{1}{n}-\frac{1}{n+1}=1-\frac{1}{n+1}$$

Check this: https://en.wikipedia.org/wiki/Telescoping_series

Answer 4

"Another way" to write things down, playing with index translation in the discrete sum:

Starting from your last equation: $$S_{n} = \sum_{k=1}^{n} \frac{1}{k} - \sum_{k=1}^{n} \frac{1}{k+1}$$ you can translate indices $k=k'+1$ in the first sum:

$$S_{n} = \sum_{k'=0}^{n-1} \frac{1}{k'+1} - \sum_{k=1}^{n} \frac{1}{k+1}$$

then make apparent the "extra" terms:

$$S_{n} = 1+ \sum_{k'=1}^{n-1} \frac{1}{k'+1} - \sum_{k=1}^{n-1} \frac{1}{k+1}-\frac{1}{n+1}$$

and you get your result: $$S_{n} = 1 -\frac{1}{n+1}$$