The value of frac{1}{(1 + a)(2 + a)} + frac{1 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The given series is $S = \sum_{n=1}^{\infty} \frac{1}{(n + a)(n + 1 + a)}$.Using the method of partial fractions,the $n^{th}$ term $T_n$ can be wr

Vedclass · Accepted Answer

$\frac{1}{1 + a}$

Vedclass · Answer

(A) The given series is $S = \sum_{n=1}^{\infty} \frac{1}{(n + a)(n + 1 + a)}$.Using the method of partial fractions,the $n^{th}$ term $T_n$ can be written as:$T_n = \frac{1}{(n + a)(n + 1 + a)} = \frac{1}{n + a} - \frac{1}{n + 1 + a}$.Now,we write the sum of the first $n$ terms $S_n$:$S_n = T_1 + T_2 + T_3 + \dots + T_n$$S_n = \left( \frac{1}{1 + a} - \frac{1}{2 + a} ight) + \left( \frac{1}{2 + a} - \frac{1}{3 + a} ight) + \left( \frac{1}{3 + a} - \frac{1}{4 + a} ight) + \dots + \left( \frac{1}{n + a} - \frac{1}{n + 1 + a} ight)$.This is a telescoping series where intermediate terms cancel out:$S_n = \frac{1}{1 + a} - \frac{1}{n + 1 + a}$.To find the sum to infinity,we take the limit as $n 	o \infty$:$S_{\infty} = \lim_{n 	o \infty} \left( \frac{1}{1 + a} - \frac{1}{n + 1 + a} ight) = \frac{1}{1 + a} - 0 = \frac{1}{1 + a}$.

The value of $\frac{1}{(1 + a)(2 + a)} + \frac{1}{(2 + a)(3 + a)} + \frac{1}{(3 + a)(4 + a)} + \dots + \infty$ is,(where $a$ is a constant).

Similar Questions

$A$ series whose $n^{th}$ term is $\left( \frac{n}{x} \right) + y$,the sum of $r$ terms will be

The sum to infinity of the following series $2 + \frac{1}{2} + \frac{1}{3} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{2^3} + \frac{1}{3^3} + \dots$ will be

The sum of all the digits of the numbers from $1$ to $100$ is

If the first term of a $G.P.$ is $729$ and $7^{th}$ term is $64,$ determine $S_{7}$.

The number of terms common between the two series $2 + 5 + 8 + \dots$ up to $50$ terms and the series $3 + 5 + 7 + 9 + \dots$ up to $60$ terms is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module