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

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

Question

(D) Let the given series be $S = 1 + 2 + \frac{1}{2!} + \frac{2}{3!} + \frac{1}{4!} + \frac{2}{5!} + \dots$We can rewrite the series by separating the

Vedclass · Accepted Answer

$\frac{3e - e^{-1}}{2}$

Vedclass · Answer

(D) Let the given series be $S = 1 + 2 + \frac{1}{2!} + \frac{2}{3!} + \frac{1}{4!} + \frac{2}{5!} + \dots$We can rewrite the series by separating the terms with $1$ in the numerator and terms with $2$ in the numerator:$S = (1 + \frac{1}{2!} + \frac{1}{4!} + \dots) + 2(1 + \frac{1}{3!} + \frac{1}{5!} + \dots)$Note that the standard series for $e$ and $e^{-1}$ are:$e = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \dots$$e^{-1} = 1 - 1 + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \dots$Thus,$\frac{e + e^{-1}}{2} = 1 + \frac{1}{2!} + \frac{1}{4!} + \dots$ and $\frac{e - e^{-1}}{2} = 1 + \frac{1}{3!} + \frac{1}{5!} + \dots$Substituting these into the expression for $S$:$S = \frac{e + e^{-1}}{2} + 2 \left( \frac{e - e^{-1}}{2} \right)$$S = \frac{e + e^{-1} + 2e - 2e^{-1}}{2} = \frac{3e - e^{-1}}{2}$.

The sum of the infinite series $1 + 2 + \frac{1}{2!} + \frac{2}{3!} + \frac{1}{4!} + \frac{2}{5!} + \dots$ is

Similar Questions

$\sum_{n=1}^{\infty} \frac{2n}{(2n+1)!}$ is equal to

In the expansion of $\frac{e^{7x} + e^{3x}}{e^{5x}}$,the constant term is

In the expansion of $(e^x - 1)^2$,the coefficient of $x^4$ will be

$\sum_{k=1}^{\infty} \frac{1}{k !} \left(\sum_{n=1}^k 2^{n-1}\right)$ is equal to

If $a = \sum\limits_{n = 0}^\infty {\frac{{{x^{3n}}}}{{(3n)!}}} ,\,b = \sum\limits_{n = 1}^\infty {\frac{{{x^{3n - 2}}}}{{(3n - 2)!}}} $ and $c = \sum\limits_{n = 1}^\infty {\frac{{{x^{3n - 1}}}}{{(3n - 1)!}}} $ then the value of ${a^3} + {b^3} + {c^3} - 3abc = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module