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)


 * Yes, probably. I've at least got a proof in my head requiring no further action when we have $a_{i,n} \to a_i$ and existence of $\displaystyle \lim_{n \to \infty} \sum_{i \mathop = 1}^n a_{i,n}$; simply estimating the error term (say for $k \ge N$, estimate independent of $n$) with inaccuracy $\epsilon/2$, then all the other terms to inaccuracy $\epsilon/2N$, so in total inaccuracy $\epsilon$, done. But a link would be good. --Lord_Farin (talk) 22:12, 17 October 2012 (UTC)