The sum of the series frac{1}{2 !} + frac{1+2 | vedclass

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

Question

(A) The $n$th term of the series is given by $T_n = \frac{1+2+3+\ldots+n}{(n+1) !}$.Using the sum of first $n$ natural numbers,$T_n = \frac{n(n+1)}{2(

Vedclass · Accepted Answer

$\frac{e}{2}$

Vedclass · Answer

(A) The $n$th term of the series is given by $T_n = \frac{1+2+3+\ldots+n}{(n+1) !}$.Using the sum of first $n$ natural numbers,$T_n = \frac{n(n+1)}{2(n+1)!} = \frac{n}{2(n)!} = \frac{1}{2(n-1)!}$.For $n=1, 2, 3, \ldots$,the terms are $T_1 = \frac{1}{2(0!)}, T_2 = \frac{1}{2(1!)}, T_3 = \frac{1}{2(2!)}, \ldots$.The sum $S = \sum_{n=1}^{\infty} T_n = \frac{1}{2} \left[ \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \ldots \right]$.Since $e = \sum_{k=0}^{\infty} \frac{1}{k!} = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \ldots$,we have $S = \frac{1}{2} e$.

The sum of the series $\frac{1}{2 !} + \frac{1+2}{3 !} + \frac{1+2+3}{4 !} + \ldots$ is equal to :

Similar Questions

$\frac{2}{2!} + \frac{2+4}{3!} + \frac{2+4+6}{4!} + \dots$ is equal to

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

$\frac{\frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + \dots \infty}{1 + \frac{1}{3!} + \frac{1}{5!} + \frac{1}{7!} + \dots \infty} = $

$1 + \frac{1 + 3}{2!} + \frac{1 + 3 + 5}{3!} + \frac{1 + 3 + 5 + 7}{4!} + \dots \infty = $

The sum of the series $1 + \frac{3}{2!} + \frac{7}{3!} + \frac{15}{4!} + \dots \text{to } \infty$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module