Summation of Factorial

Calculus Level 2

$\large\displaystyle\sum _{n=1}^{\infty}{((n-1)!)^{-1}} =\ ?$

e

\phi

\pi

\infty

1 solution

Samrit Pramanik
Apr 29, 2015

We have, $e^x = \frac{x^0}{0!} + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + ...$

So, putting $x=1$ , we get

$e = \large\displaystyle\sum_{n=1}^{\infty}{((n-1)!)^{-1}}$

So the required answer is $e$

This is trivial if one knows the Maclaurin series of $e^x$ . Since the problem asks such a question, a proper solution should show the derivation.

Hint: Use Taylor's Theorem .

Prasun Biswas - 5 years, 11 months ago