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Prove that

$${e^{-1} =2(\frac{1}{3!} + \frac{2}{5!} + \frac{3}{7!} + \frac{4}{9!} ....)}$$.

I am unable to solve it. I know I have to solve it using expansion of ${e^x}$.But I am unable to understand the algebraic manipulation that I have to perform to solve it. Please help me. Thank you! :))

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What are the next terms? – Did May 3 '14 at 13:56
Giving just 2 terms isn't enough, please add a few more or write a formula. – user88595 May 3 '14 at 13:57
I think it may be $\frac{r}{(2r+1)!}$ – evil999man May 3 '14 at 13:59
Did you miss a + sign? – evil999man May 3 '14 at 13:59

1 Answer 1

up vote 12 down vote accepted

$$\begin{align}e^{-1}&=\sum_{n=0}^\infty\frac{(-1)^n}{n!} \\&=\sum_{n=0}^\infty\left(\frac{(-1)^{2n}}{(2n)!}+\frac{(-1)^{2n+1}}{(2n+1)!}\right)\\&=\sum_{n=0}^\infty\frac{(2n+1)-1}{(2n+1)!}\\&=2\sum_{n=0}^\infty\frac{n}{(2n+1)!}\\&=2\sum_{n=1}^\infty\frac{n}{(2n+1)!}\end{align}$$

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Two bottom expressions are identical. Did you mean to have $2$ inside the sum in the pre-last one? – Ruslan May 3 '14 at 14:49
@Ruslan Not identical. $n=0\to\infty$ in the first, but $n=1\to\infty$ in the second. – Justin May 3 '14 at 16:27
Ah, didn't notice this. – Ruslan May 3 '14 at 16:27

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