frac{2}{3!} + frac{4}{5!} + frac{6}{7!} + dot | vedclass

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

Question

(D) Let the series be $S = \sum_{n=1}^{\infty} \frac{2n}{(2n+1)!}$.The general term is $T_n = \frac{2n}{(2n+1)!} = \frac{(2n+1)-1}{(2n+1)!} = \frac{1}

Vedclass · Accepted Answer

$1/e$

Vedclass · Answer

(D) Let the series be $S = \sum_{n=1}^{\infty} \frac{2n}{(2n+1)!}$.The general term is $T_n = \frac{2n}{(2n+1)!} = \frac{(2n+1)-1}{(2n+1)!} = \frac{1}{(2n)!} - \frac{1}{(2n+1)!}$.Summing from $n=1$ to $\infty$:$S = \left( \frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + \dots \right) - \left( \frac{1}{3!} + \frac{1}{5!} + \frac{1}{7!} + \dots \right)$.We know that $e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots$ and $e^{-1} = 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \dots$.Thus,$\frac{e+e^{-1}}{2} = 1 + \frac{1}{2!} + \frac{1}{4!} + \dots \implies \frac{1}{2!} + \frac{1}{4!} + \dots = \frac{e+e^{-1}}{2} - 1$.And $\frac{e-e^{-1}}{2} = \frac{1}{1!} + \frac{1}{3!} + \frac{1}{5!} + \dots \implies \frac{1}{3!} + \frac{1}{5!} + \dots = \frac{e-e^{-1}}{2} - 1$.Substituting these values:$S = \left( \frac{e+e^{-1}}{2} - 1 \right) - \left( \frac{e-e^{-1}}{2} - 1 \right) = \frac{e+e^{-1}-e+e^{-1}}{2} = e^{-1} = \frac{1}{e}$.

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

Similar Questions

The money invested in a company is compounded continuously. ₹ $400$ invested today becomes ₹ $800$ in $6$ years. What will it become at the end of $33$ years? (Given $\sqrt{2} \approx 1.4142$)

$3 + \frac{5}{1!} + \frac{7}{2!} + \frac{9}{3!} + \dots \infty = $

In the expansion of $\frac{a + bx}{e^x}$,the coefficient of $x^r$ is

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

In the expansion of $\frac{e^{5x} + e^x}{e^{3x}}$,the coefficient of $x^4$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module