$\sum_{n=0}^\infty \frac{1}{(2n)!}=\cosh(1)$

Question

Is there any way to show that $$\sum_{n=0}^\infty \frac{1}{(2n)!}=\cosh(1) = \frac{1+e^2}{2e}$$ knowing that $$\sum_{n=0}^\infty \frac{1}{n!}=e$$ I guess that we can also show that $\sum_{n=0}^\infty \frac{1}{(2n-1)!}=\sinh(1) $ but I have no clue of how to get rid of the "2" in the factorial...

Answer 1

No, the fact that $\sum_{n=0}^\infty \frac 1{n!} = e$ is insufficient (by any reasonable interpretation of what it means to "use a fact"). However, you could reach your result if you use the additional fact that $$ \sum_{n=0}^\infty \frac{(-1)^n}{n!} = e^{-1}. $$ In particular, we can note that $$ \sum_{n=0}^\infty \frac{1}{(2n)!} = \frac 12 \left( \sum_{n=0}^\infty \frac{1}{n!} + \sum_{n=0}^\infty \frac{(-1)^n}{n!} \right) $$

Answer 2

Write down the series expansions for $e^x$ and $e^{-x}$ around $0$. Add the two series and divide by $2$. Similarly, you can subtract them, to get $\sinh$. $$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}\\e^{-x}=\sum_{n=0}^\infty\frac{(-1)^nx^n}{n!}$$ When you add these two expressions, the odd terms will cancel. If you subtract them, the even terms will cancel out.

Answer 3

Let $\sum_{n=0}^\infty \frac{1}{n!} = e$, and $\sum_{n=0}^\infty \frac{(-1)^n}{n!} = x$. I claim that $x = 1/e$. Since the series converge absolutely, we can rearrange

$$ \eqalign{e x &= \sum_{n=0}^\infty \sum_{m=0}^\infty \frac{1}{n!} \frac{(-1)^m}{m!}\cr &= \sum_{k=0}^\infty \sum_{n=0}^k \frac{1}{n!} \frac{(-1)^{k-n}}{(k-n)!}\cr &= \sum_{k=0}^\infty \frac{1}{k!} \sum_{n=0}^k {k \choose n} (-1)^{k-n}\cr &= \sum_{k=0}^\infty \frac{1}{k!} (1-1)^k}$$ where the last equality uses the binomial theorem. But $(1-1)^k = 0^k = 1$ for $k=0$ (check the sum explicitly for the case $k=0$ if you're uncomfortable with this) and $0$ for $k > 0$, so we conclude that $ex = 1$.

And now, as others have mentioned, $$\sum_{n=0}^\infty \frac{1}{(2n)!} = \frac{1}{2} \sum_{n=0}^\infty \frac{1 + (-1)^n}{n!} = \frac{e + 1/e}{2} = \cosh(1)$$