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

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

Question

(B) We know the expansion of $\cosh(x) = \frac{e^x + e^{-x}}{2} = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!} + \dots \infty$. Setting $x = 1

Vedclass · Accepted Answer

$\frac{(e - 1)^2}{2e}$

Vedclass · Answer

(B) We know the expansion of $\cosh(x) = \frac{e^x + e^{-x}}{2} = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!} + \dots \infty$. Setting $x = 1$,we get: $\frac{e^1 + e^{-1}}{2} = 1 + \frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + \dots \infty$. $\frac{1}{2} \left( e + \frac{1}{e} \right) = 1 + \frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + \dots \infty$. Therefore,$\frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + \dots \infty = \frac{1}{2} \left( \frac{e^2 + 1}{e} \right) - 1$. $= \frac{e^2 + 1 - 2e}{2e} = \frac{(e - 1)^2}{2e}$.

Find the sum of the series $\frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + \dots \infty$.

Similar Questions

Find the sum to infinity of the series $\frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \dots$

If $x=1+\frac{1}{2 \times 1 !}+\frac{1}{4 \times 2 !}+\frac{1}{8 \times 3 !}+\ldots$ and $y=1+\frac{x^{2}}{1 !}+\frac{x^{4}}{2 !}+\frac{x^{6}}{3 !}+\ldots$, then the value of $\log_{e} y$ is

The coefficient of $x^n$ in the expansion of $\frac{e^{7x} + e^x}{e^{3x}}$ is

$\left( {1 + \frac{1}{{2!}} + \frac{1}{{4!}} + \dots} \right) \left( {1 + \frac{1}{{3!}} + \frac{1}{{5!}} + \dots} \right) = $

The sum of the series $\frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \dots$ up to infinity is

Vedclass Test Series

Exam Paper Generator

Online Exam Module