I know that $\displaystyle e$ can be approximated by the form $\displaystyle 1 + \frac {1}{1!} + \frac {2}{2!} + \frac {3}{3!} + \frac {4}{4!} +... \frac {n}{n!}$ .

Would $\displaystyle \frac {2^1}{1!} + \frac {2^2}{2!} + \frac {2^3}{3!} + \frac {2^4}{4!} +... \frac {2^n}{n!} = e^2$ ?

And how would you determine an aproximate value for $\displaystyle \frac {a^1}{1!} + \frac {a^2}{2!} + \frac {a^3}{3!} + \frac {a^4}{4!} +... \frac {a^n}{n!}$ ?