Determining: $\lim_{n\rightarrow \infty}e^{-n}\sum_{k=n}^\infty n^k/k!$

Question

What is the value of $\lim_{n\rightarrow \infty}e^{-n}\sum_{k=n}^\infty n^k/k!$ ?

I have tried initially but could not proceed any further. What I have tried is:

$$\lim_{n\rightarrow \infty}e^{-n}\sum_{k=n}^\infty{n^k \over k!}\\ =\lim_{n\rightarrow \infty}e^{-n}\left[e^n-\sum_{k=0}^{n-1}{n^k \over k!}\right]$$

I got no clue after this. I am not sure if how that limit can be be determined.

Any kind of help will be welcome.

$1+n+n^2/2!+....n^{n-1}/(n-1)!\rightarrow e^n$, so the limit is $0$. Where are you having a problem in that? — QED, Dec 30 '17 at 12:24
How did you write $lim_{n\rightarrow \infty}\sum_{k=n}^\infty n^k/k!\ =\lim_{n\rightarrow \infty}[e^n-(1+n+n^2/2!+....n^{n-1}/(n-1)!)]$ — QED, Dec 30 '17 at 12:34
Because $e^n=1 + n + n^2 +....+n^{n-1}/(n-1)! + n^n/n! +......$ — Sourav Sarkar, Dec 30 '17 at 12:41
Yeah... It is correct... That is what I pointed out that $e^n=\sum_{k=1}^{\infty}n^k/k!$, he has already used that, but he is aking me how I knew that. — QED, Dec 30 '17 at 12:41
@daniel Fisher this is not 100% a duplicate . here we have squeed sum — Guy Fsone, Dec 30 '17 at 13:07
@GuyFsone $\lim_{n\to\infty} (1 - a_n + O(n^{-1/2}))$ is close enough to $\lim_{n\to\infty} a_n$. — Daniel Fischer, Dec 30 '17 at 13:09
@DanielFischer also see that there is a term $\frac{e^{-n}n^n}{n!}$ which require Stirling formula to handle it. — Guy Fsone, Dec 30 '17 at 13:22
@GuyFsone So what? You need far heavier tools than Stirling's formula to determine the limit anyway. — Daniel Fischer, Dec 30 '17 at 13:23
If the $\lim_{n \rightarrow \infty}e^{-n}\sum_{k=0}^n n^k/k! =1/2$ , then i guess this limit value will be 1/2 too. — Sourav Sarkar, Dec 30 '17 at 13:36
Right, @SouravSarkar. Apart from the term $\frac{n^n}{e^n\cdot n!}$, which tends to $0$ by Stirling's approximation, your sequence is $1-e^{-n}\sum_{k = 0}^n \frac{n^k}{k!}$. — Daniel Fischer, Dec 30 '17 at 13:41
Why the votes to reopen? From the work of the OP, the rest of the question may easily be answered by the duplicate link. — Simply Beautiful Art, Dec 30 '17 at 14:20
@SimplyBeautifulArt why not? just close it as soon as it is open . I voted for reopen — Guy Fsone, Dec 30 '17 at 14:24
@GuyFsone I merely would like to know some reason that 5 users decided to reopen the question, because surely my own judgement may not be better than 5 others? What wrong is there in asking? — Simply Beautiful Art, Dec 30 '17 at 14:51

score 0 · Answer 1 · answered Dec 30 '17 at 14:49

we have $$e^{-n}\sum_{k=n}^\infty \frac{n^k}{k!} =e^{-n}\left[e^n-\sum_{k=0}^{n-1} \frac{n^k}{k!}\right]= \left[1+ \frac{e^{-n}n^n}{n!}-e^{-n}\sum_{k=0}^{n} \frac{n^k}{k!}\right] $$ By Stirling formula $$ \frac{e^{-n}n^n}{n!}\sim \frac{1}{\sqrt{2n\pi}}\to0$$ from and from here:Evaluating $\lim\limits_{n\to\infty} e^{-n} \sum\limits_{k=0}^{n} \frac{n^k}{k!}$ we have $$\lim_{n\rightarrow \infty}\left[e^{-n}\sum_{k=0}^{n} \frac{n^k}{k!}\right] = \frac12$$ thus $$\lim_{n\rightarrow \infty}e^{-n}\sum_{k=n}^\infty \frac{n^k}{k!}=1-\frac12$$

Determining: $\lim_{n\rightarrow \infty}e^{-n}\sum_{k=n}^\infty n^k/k!$

1 Answers1