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can you please tell me if the folowing equation is correct, and if it is, how am I proove this?

$$\lim_{n\rightarrow\infty}\sum_{k=0}^n\frac{n!}{k!(n-k)!}\cdot\frac{1}{n^k}=\sum_{k=0}^\infty\lim_{n\rightarrow\infty}\frac{n!}{k!(n-k)!}\cdot\frac{1}{n^k}$$

Thanks

tzur dolev
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    See https://math.stackexchange.com/questions/23057/under-what-condition-we-can-interchange-order-of-a-limit-and-a-summation – Joe Jul 09 '20 at 10:12
  • It is, as the limit of the sum of two quantities is equal to the sum of their limits – Dhanvi Sreenivasan Jul 09 '20 at 10:19
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    @Dhanvi Sreenivasan, you cannot simply infer from that fact about a finite sum that the limits can be exchanged for an infinite sum. See the link. – Joe Jul 09 '20 at 10:23
  • @Joe Additional complication compared to your linked question, as the sum on the left-hand side isn't infinite, it has $n$ terms. – Arthur Jul 09 '20 at 10:27
  • Oh, I see. Thank you! – Joe Jul 09 '20 at 10:29
  • Could you consider both sides to be infinite sums by multiplying by an indicator function $I(k \le n)$? – Joe Jul 09 '20 at 10:33
  • In general, swapping the limit and the sum like this isn't valid, but in this particular case it turns out to be true (it's the pivotal step in showing that $$\lim_{n\to\infty}\left(1+\frac1n\right)^n=\sum_{k=0}^\infty \frac 1{k!}$$which is a well-known identity). I don't remember the formal proof, though. – Arthur Jul 09 '20 at 10:37
  • @Arthur . Given $\epsilon >0,$ take $n_1$ large enough that $\sum_{n>n_1}1/n!<\epsilon/2.$ Take $n_2>n_1,$ large enough that $1/k!-n^{-k} \binom {n}{k}<\epsilon/2(n_1+1)$ whenever $k\le n_1$ and $n>n_2.$ – DanielWainfleet Jul 09 '20 at 16:31
  • You are merely lucky as it is not, in general, a logical move. – DanielWainfleet Jul 09 '20 at 16:38

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