frac{1}{2!} + frac{1 + 2}{3!} + frac{1 + 2 + | vedclass

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

Question

(C) The general term of the series is $T_n = \frac{1 + 2 + \dots + n}{(n + 1)!}$.Using the sum of first $n$ natural numbers,$T_n = \frac{n(n + 1)}{2(n

Vedclass · Accepted Answer

$e/2$

Vedclass · Answer

(C) The general term of the series is $T_n = \frac{1 + 2 + \dots + n}{(n + 1)!}$.Using the sum of first $n$ natural numbers,$T_n = \frac{n(n + 1)}{2(n + 1)!} = \frac{n}{2(n + 1)!}$.To simplify,we write $T_n = \frac{(n + 1) - 1}{2(n + 1)!} = \frac{1}{2} \left( \frac{1}{n!} - \frac{1}{(n + 1)!} \right)$.Summing from $n = 1$ to $\infty$,$S = \sum_{n=1}^{\infty} \frac{1}{2} \left( \frac{1}{n!} - \frac{1}{(n + 1)!} \right) = \frac{1}{2} \left[ (\frac{1}{1!} - \frac{1}{2!}) + (\frac{1}{2!} - \frac{1}{3!}) + \dots \right]$.This is a telescoping series,so $S = \frac{1}{2} (1) = \frac{1}{2}$.

$\frac{1}{2!} + \frac{1 + 2}{3!} + \frac{1 + 2 + 3}{4!} + \dots \infty = $

Similar Questions

The sum of the series $\sum_{n=1}^{\infty} \frac{n^{2}+6 n+10}{(2 n+1) !}$ is equal to :

$1 + \frac{3}{1!} + \frac{5}{2!} + \frac{7}{3!} + ....\infty = $

$1.5 + \frac{2.6}{1!} + \frac{3.7}{2!} + \frac{4.8}{3!} + \dots$ is equal to

The sum of the series $\frac{4}{1!} + \frac{11}{2!} + \frac{22}{3!} + \frac{37}{4!} + \frac{56}{5!} + \dots$ is

$1 + \frac{1 + 2}{1!} + \frac{1 + 2 + 3}{2!} + \frac{1 + 2 + 3 + 4}{3!} + \dots \infty = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module