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I need to find

$$I=\lim_{k\to\infty}\left(\sum _{n=0}^kn!\ ^kC_n \left(\frac{3}{k}\right)^{n}-\dfrac{3^kk!e^{\tfrac k3}}{k^k} \right).$$

This was because I got stuck during an attempt to solve this question by methods accessible to a high-schooler. This is basically the error in the approximation of the right hand term by the sum, by this post. Any help is appreciated!

Answer is supposed to be

$$\dfrac 12$$

P.S. : If you want to see “my effort” then see my answer to the linked post. :)

insipidintegrator
  • 6,730
  • And this series represents... – FShrike Jul 29 '22 at 18:18
  • What do you mean…pls pls elaborate…not an AGP, not a GP, not(?) a Riemann sum… – insipidintegrator Jul 29 '22 at 18:30
  • I'm just asking what that limit is supposed to mean. Yes, you linked your post, but there really isn't much context – FShrike Jul 29 '22 at 18:32
  • The series represents one-third of the limit of integral $\lim \limits_{k \to \infty} \int_{0}^{k} \left(1-\frac{x}{k}\right)^k \cdot e^{\frac{x}{3}} dx$ . I hope you have seen my attempt. Unless I have made some crass mistake, this should be it… – insipidintegrator Jul 29 '22 at 18:36
  • @FShrike I don’t know what other context to add… – insipidintegrator Jul 29 '22 at 18:42
  • The answer you linked reduces the problem to finding $$\lim_{k \to \infty} \frac{k!}{(k/3)^k} \left(e^{k/3} - \sum_{n=0}^{k}{\frac{(k/3)^n}{n!}}\right)$$ It should be possible to adapt an answer from this post, after which you'd be done by Stirling's approximation. – Ant Jul 29 '22 at 20:29

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