Find the sum to n terms of the series frac{1} | vedclass

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

Question

(A) The given series is $\frac{1}{1 	imes 2} + \frac{1}{2 	imes 3} + \frac{1}{3 	imes 4} + \ldots$The $n^{th}$ term is $a_{n} = \frac{1}{n(n+1)}$.U

Vedclass · Accepted Answer

$\frac{n}{n+1}$

Vedclass · Answer

(A) The given series is $\frac{1}{1 	imes 2} + \frac{1}{2 	imes 3} + \frac{1}{3 	imes 4} + \ldots$The $n^{th}$ term is $a_{n} = \frac{1}{n(n+1)}$.Using partial fractions,we can write $a_{n} = \frac{1}{n} - \frac{1}{n+1}$.Now,the sum of $n$ terms is $S_{n} = \sum_{k=1}^{n} a_{k} = \sum_{k=1}^{n} \left( \frac{1}{k} - \frac{1}{k+1} ight)$.Expanding the sum:$S_{n} = \left( \frac{1}{1} - \frac{1}{2} ight) + \left( \frac{1}{2} - \frac{1}{3} ight) + \left( \frac{1}{3} - \frac{1}{4} ight) + \ldots + \left( \frac{1}{n} - \frac{1}{n+1} ight)$.This is a telescoping series where intermediate terms cancel out:$S_{n} = 1 - \frac{1}{n+1}$.Simplifying the expression:$S_{n} = \frac{n+1-1}{n+1} = \frac{n}{n+1}$.

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

Similar Questions

The sum of the infinite series $\cot ^{-1}\left(\frac{7}{4}\right)+\cot ^{-1}\left(\frac{19}{4}\right)+\cot ^{-1}\left(\frac{39}{4}\right)+\cot ^{-1}\left(\frac{67}{4}\right)+\ldots \ldots$ is :-

The value of $\lim _{n \rightarrow \infty} \left( \sum_{k=1}^n \frac{k^3+6 k^2+11 k+5}{(k+3)!} \right)$ is:

The sum $\sum\limits_{r = 1}^{10} {({r^2} + 1) \times r!}$ is equal to

Find the sum of the series $1 \cdot 2 \cdot 3 + 2 \cdot 3 \cdot 4 + 3 \cdot 4 \cdot 5 + \dots$ up to $n$ terms.

If $\sum_{r=1}^{25} \left( \frac{r}{r^{4}+r^{2}+1} \right) = \frac{p}{q}$ where $p$ and $q$ are positive integers such that $\gcd(p,q)=1$, then $p+q$ is equal to . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module