I am trying to show that as $\bar{n} \rightarrow \infty$, the following term tends to $0$ (at least this is what I think). $$ \sum_{n=\bar{n}}^\infty \frac{\bar{n}^n}{n!}. $$ Since the summand is $$e^{\bar{n}} - \sum_{n=0}^{\bar{n}-1} \frac{\bar{n}^n}{n!},$$ it seems we have to show that the lagrange remainder tends to $0$, but here $\bar{n}$ has a limit. Am I going in the wrong direction?
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1See https://math.stackexchange.com/q/160248/42969: $\sum_{n=k}^\infty \frac{k^n}{n!} \sim \frac 12 e^k$. – Martin R Jun 16 '24 at 14:36
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@MartinR, so I am looking at a divergent series, quite against my intuition. – Cherryblossoms Jun 16 '24 at 14:40
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2Btw, the use of $n$ and $\bar{n}$ as the two variables makes this difficult to read. – Martin R Jun 16 '24 at 14:40