Prove the following expansion.

Question

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! :))

Giving just 2 terms isn't enough, please add a few more or write a formula. — user88595, May 03 '14 at 13:57
Related : http://math.stackexchange.com/questions/221951/evaluate-sum-k-1-infty-frack2k-1, http://math.stackexchange.com/questions/581603/evaluate-frac13-frac14-frac12-frac15-frac13-dots and http://math.stackexchange.com/questions/348061/exponential-and-logarithmic-series-find-the-sum-of-22-32-242-3-t — lab bhattacharjee, May 03 '14 at 15:40

score 12 · Accepted Answer · answered May 03 '14 at 14:00

12

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

answered May 03 '14 at 14:00

Hagen von Eitzen

382,479

Two bottom expressions are identical. Did you mean to have $2$ inside the sum in the pre-last one? – Ruslan May 03 '14 at 14:49
1

@Ruslan Not identical. $n=0\to\infty$ in the first, but $n=1\to\infty$ in the second. – Justin May 03 '14 at 16:27
Ah, didn't notice this. – Ruslan May 03 '14 at 16:27

Prove the following expansion.

1 Answers1