# A infinite series expansion for $e^e$.

How can $e^e$ be expressed in an infinite series with as much simplification as possible.

• I wrote the series of $e^x$ by keeping $x$ as $e$ and from there I also expanded every $e$ in this expansion now I was thing about expanding it further by binomial theorem but I am not able to understand how can i use binomial theorem here and how much can this be simplified in another words I am trying to write this series as simple as expansion of $e$ , is it possible and how it can be done.

• Any help will be highly appreciated , thanks in advance.

• Have a look at the Wikipedia article Bell Number, particularly the intro and section 3.2 (Generating function). – DanielWainfleet May 28 '17 at 15:05

Let $f(x)=e^x$ Then $$f^{(1)}(x)=f^{(2)}(x)=...=f^{(n)}(x)=...=e^x, \forall x\in\Bbb R.$$ Then applying Taylor's theorem we get - $$f(x)=f(0)+xf^{(1)}(0)+\frac {x^2}{2!} f^{(2)}(0)+...+\frac {x^n}{n!} f^{(n)}(\phi), 0\lt\phi\lt1.$$ Then you will get $$e^x=1+x+\frac{x^2}{2!}+...$$ i.e. $$e^x=\sum_{n=0}^{\infty}\frac {x^n}{n!}$$ And hence $$e^e=\sum_{n=0}^{\infty}\frac {e^n}{n!}$$ $$=\sum_{n=0}^{\infty}\frac{\sum_{i=0}^{\infty}\frac{n^i}{i!}}{n!}.$$
• Quoting the OP: "I wrote the series of $e^x$ by keeping $x$ as $e$". – Shaun May 28 '17 at 12:54
$$e^e=\sum_{k\ge 0}\frac{e^k}{k!}=\sum_{k,\,l\ge 0}\frac{k^l}{k!l!}$$