frac{frac{1}{2!} + frac{1}{4!} + frac{1}{6!} | vedclass

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(B) We know that the exponential series are given by:$e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \dots \infty$$e^{-1} = 1 - \

Vedclass · Accepted Answer

$\frac{e - 1}{e + 1}$

Vedclass · Answer

(B) We know that the exponential series are given by:$e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \dots \infty$$e^{-1} = 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \dots \infty$Adding these,we get $e + e^{-1} = 2(1 + \frac{1}{2!} + \frac{1}{4!} + \dots \infty)$,so $\frac{1}{2!} + \frac{1}{4!} + \dots \infty = \frac{e + e^{-1} - 2}{2}$.Subtracting these,we get $e - e^{-1} = 2(\frac{1}{1!} + \frac{1}{3!} + \frac{1}{5!} + \dots \infty)$,so $1 + \frac{1}{3!} + \frac{1}{5!} + \dots \infty = \frac{e - e^{-1}}{2}$.Substituting these into the expression:$\frac{\frac{e + e^{-1} - 2}{2}}{\frac{e - e^{-1}}{2}} = \frac{e + \frac{1}{e} - 2}{e - \frac{1}{e}} = \frac{\frac{e^2 + 1 - 2e}{e}}{\frac{e^2 - 1}{e}} = \frac{(e - 1)^2}{(e - 1)(e + 1)} = \frac{e - 1}{e + 1}$.

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

Similar Questions

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

$(1 + 3)\log_e 3 + \frac{1 + 3^2}{2!} (\log_e 3)^2 + \frac{1 + 3^3}{3!} (\log_e 3)^3 + \dots \infty = $

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

$1 + x \log_e a + \frac{x^2}{2!} (\log_e a)^2 + \frac{x^3}{3!} (\log_e a)^3 + \dots = $

If $S_n$ denotes the sum of the products of the first $n$ natural numbers taken two at a time,then $\sum\limits_{n = 0}^\infty \frac{S_n}{(n + 1)!} = $

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