Show that $$\sum_{k\,=\,1}^{\infty }\frac{1}{k(k+1)}=\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3 }+\frac{1}{3\cdot 4}+\cdots =\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\cdots =1$$

Solution We begin with the sequence $\{S_{n}\}$ of partial sums, $$\begin{array}{cl} S_{1} &= \dfrac{1}{1\cdot 2} \\[10pt] S_{2} &= \dfrac{1}{1\cdot 2}+\dfrac{1}{2\cdot 3} \\[10pt] S_{3} &= \dfrac{1}{1\cdot 2}+\dfrac{1}{2\cdot 3}+\dfrac{1}{3\cdot 4} \\ \vdots & \\ S_{n} &= \dfrac{1}{1\cdot 2}+\dfrac{1}{2\cdot 3}+\dfrac{1}{3\cdot 4}+\cdots +\dfrac{1}{n(n+1)} \\ \vdots & \end{array}$$

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Since $$\begin{array}{l@{\hspace{14.5pt}}l} \dfrac{1}{n(n+1)}=\dfrac{1}{n}-\dfrac{1}{n+1} & {\color{#0066A7}{{\hbox{Use partial fractions.}}}} \end{array}$$

$S_{n}$ can be written as $$S_{n}=\left( {\frac{1}{1}-\frac{1}{2}}\right) +\left( {\frac{1}{2}-\frac{1}{3 }}\right) +\cdots +\left( {\frac{1}{n-1}-\frac{1}{n}}\right) +\left( {\frac{1 }{n}-\frac{1}{n+1}}\right)$$

After removing parentheses notice that all the terms except the first and last cancel, so that $$S_{n}=1-\frac{1}{n+1}$$

Then $$\lim_{n\rightarrow \infty }S_{n}=\lim_{n\rightarrow \infty }\left( {1-\frac{1 }{n+1}}\right) =1$$

The series $\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{k(k+1)}$ converges, and its sum is $1$.