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

Asked by Abhisek | 1 year ago |  64

##### Solution :-

$$\dfrac{1}{1\times 2}+\dfrac{1}{2\times 3}+\dfrac{1}{3\times 4}+....$$ is the given series.

The nth term of the series is

$$a_n=\dfrac{1}{n(n+1)}$$

By partial fractions the above equation can be written as

$$a_n=\dfrac{1}{n}-\dfrac{1}{n+1}$$

Therefore,

$$a_1=\dfrac{1}{1}-\dfrac{1}{2}$$

$$a_2=\dfrac{1}{2}-\dfrac{1}{3}$$

$$a_3=\dfrac{1}{3}-\dfrac{1}{4}$$

$$a_n=\dfrac{1}{n}-\dfrac{1}{n+1}$$

$$a_1+a_2+a_3+.......+a_n$$

$$[ \dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+.....\dfrac{1}{n}]$$

$$[ \dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+.....\dfrac{1}{n+1}]$$

The sum of n terms of a series is given by the equation

$$S_n= \displaystyle\sum_{k=1}^{n}a_k$$

The sum of n terms of the given series is

$$S_n= 1-\dfrac{1}{n-1}$$

$$\dfrac{n+1-1}{n+1}=\dfrac{n}{n-1}$$

Therefore, the sum of n terms of the series is $$=\dfrac{n}{n-1}$$

Answered by Pragya Singh | 1 year ago

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