Find the sum to n terms of the series


Find the sum to n terms of the series  $\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+\ldots$


The given series is $\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+\ldots$

$n^{\text {th }}$ term, $a_{n}=\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$ (By partial fractions)



$a_{3}=\frac{1}{3}-\frac{1}{4} \ldots$


Adding the above terms column wise, we obtain

$a_{1}+a_{2}+\ldots+a_{n}=\left[\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\ldots \frac{1}{n}\right]-\left[\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\ldots \frac{1}{n+1}\right]$

$\therefore S_{n}=1-\frac{1}{n+1}=\frac{n+1-1}{n+1}=\frac{n}{n+1}$

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