The value of frac{2}{3!} + frac{4}{5!} + frac | vedclass

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(B) The general term of the series is $t_n = \frac{2n}{(2n+1)!}$.We can rewrite this as $t_n = \frac{(2n+1) - 1}{(2n+1)!} = \frac{1}{(2n)!} - \frac{1}

Vedclass · Accepted Answer

$e^{-1}$

Vedclass · Answer

(B) The general term of the series is $t_n = \frac{2n}{(2n+1)!}$.We can rewrite this as $t_n = \frac{(2n+1) - 1}{(2n+1)!} = \frac{1}{(2n)!} - \frac{1}{(2n+1)!}$.Summing from $n=1$ to $\infty$, we get $\sum_{n=1}^{\infty} \left( \frac{1}{(2n)!} - \frac{1}{(2n+1)!} \right) = \left( \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} + \dots \right)$.Recall the expansion $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$. For $x = -1$, $e^{-1} = 1 - 1 + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} + \dots$.Thus, the sum is $e^{-1} = \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} + \dots$.Therefore, the value is $e^{-1}$.

The value of $\frac{2}{3!} + \frac{4}{5!} + \frac{6}{7!} + \dots$ is

Similar Questions

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

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$\frac{1}{1!} + \frac{4}{2!} + \frac{7}{3!} + \frac{10}{4!} + \dots \infty = $

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

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