3
$\begingroup$

How is $$\sum_{k=0}^{\infty}\frac{e^{-1}}{(2k)!}=\frac{1+e^{-2}}{2}$$

Well, came upon this doing statistics, found out I didn't know why, brought the matter here. Answer much appreciated. I'm aware of geometric progressions, taylor expansions btw..

$\endgroup$
  • 4
    $\begingroup$ Rewrite it as $$\sum_{k=0}^\infty \frac{1}{(2k)!} = \frac{e+e^{-1}}2$$ $\endgroup$ – Thomas Andrews Nov 12 '15 at 21:09
  • $\begingroup$ Brilliant. ${}{}{}$ $\endgroup$ – Jerry West Nov 12 '15 at 21:16
1
$\begingroup$

$$ \sum_{k=0}^\infty \frac{e^{-1}}{(2k)!} = \frac{1}{e} \sum_{k=0}^\infty \frac{1}{(2k)!} $$ and rememeber that $$ e = \sum \frac{1}{k!} \text{ and } \frac{1}{e} = \sum \frac{(-1)^k}{k!} $$ so in the series for $e+1/e$ all odd terms cancel out and all even terms are doubled up

$\endgroup$
3
$\begingroup$

$$\sum_{k=0}^{\infty}\frac{x^{2n}}{(2k)!}=\cosh x=\frac{e^x+e^{-x}}{2}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.