# A series limit

• April 6th 2009, 09:35 AM
Moo
A series limit
Hi !

Prove that :

$\lim_{n \to \infty} ~ e^{-n} \sum_{k=0}^n \frac{n^k}{k!}=\frac 12$

:D
• April 6th 2009, 10:10 AM
NonCommAlg
Quote:

Originally Posted by Moo
Hi !

Prove that :

$\lim_{n \to \infty} ~ e^{-n} \sum_{k=0}^n \frac{n^k}{k!}=\frac 12$

:D

i think the limit originally comes from probability, something that i'm a complete idiot at it! (Rofl) there's the idea of the proof in this thread. i know a calculus proof but it's long and very ugly!

so i think it's a good challenge finding a neat calculus (analysis) solution for this question.
• April 6th 2009, 10:29 AM
Moo
Quote:

Originally Posted by NonCommAlg
i think the limit originally comes from probability, something that i'm a complete idiot at it! (Rofl) there's the idea of the proof in this thread. i know a calculus proof but it's long and very ugly!

so i think it's a good challenge finding a neat calculus (analysis) solution for this question.

Yes, that was using a probability proof :D
Same type as the previous ones :P

Laurent in the thread gave the way to solve it, I'll detail it here :)

Consider a sequence $(X_n)_{n \geq 1}$ of independent random variables, following a Poisson distribution with parameter 1 :
$\mathbb{P}(X_i=k)=e^{-1} \cdot \frac{1}{k!}$

Then let's consider $S_n=X_1+\dots+X_n$
It is easy to prove that it follows a Poisson distribution with parameter n :
$\mathbb{P}(S_n=k)=e^{-n} \cdot \frac{n^k}{k!}$

We can see that :
$\mathbb{P}(S_n \leq n)=\sum_{k=0}^n e^{-n} \cdot \frac{n^k}{k!}$
But $\{S_n \leq n\}=\{S_n-n\leq 0\}=\left\{\frac{S_n-n}{\sqrt{n}} \leq 0\right\}$

Hence $\mathbb{P}\left(\frac{S_n-n}{\sqrt{n}} \leq 0\right)=e^{-n} \sum_{k=0}^n \frac{n^k}{k!}$

Then we must check the conditions for applying the Central Limit Theorem :
$\mathbb{V}\text{ar}(X_i)=\mathbb{E}(X_i)=1<\infty \Rightarrow X_i \in L^2$
And the $X_i$ are independent and identically distributed.
The Central Limit Theorem says that :
$\frac{S_n-n \mathbb{E}(X_i)}{\sqrt{n}}=\frac{S_n-n}{\sqrt{n}}$ converges to the standard normal distribution $\mathcal{N}(0,1)$

In particular, there is a convergence of the cumulative density functions :
$\mathbb{P}\left(\frac{S_n-n}{\sqrt{n}} \leq 0\right) \longrightarrow \int_{-\infty}^0 \frac{1}{\sqrt{2\pi}} e^{-t^2/2} ~dt$, which is $\frac 12$ because the integrand is even.

This finishes the proof :
$e^{-n} \sum_{k=0}^n \frac{n^k}{k!} \longrightarrow \frac 12$

Haaa... I really like these kinds of limits that use probability, it's just great (Tongueout)

Now, if you have time, I'd be interested in seeing this proof you have that uses calculus ^^
• April 6th 2009, 11:25 AM
NonCommAlg
sure, if i basically can find the solution i had because it goes back to quite long time ago when i was a calculus I student! haha