The sum to infinity of the following series f | vedclass

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

Question

(B) The given series is $S = \sum_{n=1}^{\infty} \frac{1}{n(n+1)}$.Using partial fractions,we can write $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) The given series is $S = \sum_{n=1}^{\infty} \frac{1}{n(n+1)}$.Using partial fractions,we can write $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$.Thus,the sum of the first $N$ terms is $S_N = \sum_{n=1}^{N} \left( \frac{1}{n} - \frac{1}{n+1} ight)$.Expanding this,we get $S_N = (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + \dots + (\frac{1}{N} - \frac{1}{N+1})$.All intermediate terms cancel out,leaving $S_N = 1 - \frac{1}{N+1}$.Taking the limit as $N 	o \infty$,we get $S = \lim_{N 	o \infty} (1 - \frac{1}{N+1}) = 1 - 0 = 1$.

The sum to infinity of the following series $\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \dots$ is

Similar Questions

If $1^2 + 2^2 + 3^2 + \dots + 2009^2 = (2009)(335)(4019)$ and $(1)(2009) + 2(2008) + 3(2007) + \dots + 2009(1) = (2009)(335)(x)$,then $x$ is equal to:

Find the sum of the series $6 + 66 + 666 + \dots$ up to $n$ terms.

Let $a_n$ be a sequence such that $a_1 = 5$ and $a_{n+1} = a_n + (n - 2)$ for all $n \in N$. Then $a_{51}$ is:

$\sum_{k=1}^{\infty} \sum_{r=0}^k \frac{1}{3^k} \binom{k}{r}$ is equal to

If $2^3+4^3+6^3+\ldots+(2n)^3=h n^2(n+1)^2$,then $h$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module