Sum of Gaps

$\displaystyle\sum_{n=0}^{\infty}\left(e- \sum_{k=0}^{n} \frac{1}{k!}\right)=(e-1)+(e-1-1)+(e-1-1-\frac12)+(e-1-1-\frac12-\frac16)+...$

$\displaystyle=(1+\frac12+\frac16+\frac1{24}+\frac1{120}+...)+(\frac12+\frac16+\frac1{24}+\frac1{120}+...)+(\frac16+\frac1{24}+\frac1{120}+...)+...$

$\displaystyle=1+\frac12\times2+\frac16\times3+\frac1{24}\times4+\frac1{120}\times5+...$

$\displaystyle=1+1+\frac12+\frac16+\frac1{24}+...=e$

Solution using generating functions (not completely rigorous):

We wish to obtain a function which generates $e - P(k)$ , where $P(k)$ is the $k$ th partial sum for $e$ . We can generate $P(k)$ by the function $\frac{e^x}{1-x}$ , since $\frac{1}{1-x}$ generates partial sums. Now our function is simply $\frac{e - e^x}{1 - x}$ , and L'Hopital's rule at $x = 1$ gives $e$ .

$\sum_{n=0}^\infty \sum_{k=n+1}^\infty \frac{1}{k!} = \frac{1}{1!} + \frac{2}{2!} + \frac{3}{3!} +... = 1 + \frac{1}{1!} + \frac{2}{2!} +... = \sum_{k=0}^\infty \frac{1}{k!}= e$