Talk:Euler's Number: Limit of Sequence implies Limit of Series

Note my comment on the last step. To do it formally, we could probably pass to a double sequence of which $(1+\dfrac1n)^n$ is the diagonal, then prove the double sequence converges to $\sum (k!)^{-1}$. Paradoxically, it may be easier. --Lord_Farin (talk) 20:44, 17 October 2012 (UTC)


 * Isn't it the case that $\lim \sum = \sum \lim$ for a convergent sequence? Thought we had that proved somewhere. Or am I missing something? --prime mover (talk) 21:53, 17 October 2012 (UTC)