# Convergence of a sum to $e^{-1/2}$?

I have this sum: $\displaystyle \sum _{k=0}^{\frac{n}{2}-m} \frac{\left(-\frac{1}{2}\right)^k}{k!} \exp \left(\frac{3 (m+k)-2 (m+k)^2}{n}\right)$. I have some suspicion that it converges to $e ^{-1/2}$ for all $m=1,2....$ when $n \rightarrow \infty$; but failed to prove that. Any help?

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That is a pretty big suspicion. Is there anything similar to this thing that converges to anything known? alternatively the a book on summation of series might already have this in it. – Arjang Sep 21 '13 at 2:22

The statement is true.

Let $S_{-1} = 0$ and $S_n$ be the partial sum $\displaystyle\;\;\sum_{k=0}^{n}\frac{1}{k!}(-\frac12)^k\;\;$ for $n \ge 0$.
Let $\alpha_{n,k}$ be the double sequence defined by

$$\alpha_{n,k} = \begin{cases} \exp\left(\frac{3(m+k)-2(m+k)^2}{n}\right), & 0 \le k \le \frac{n}{2} - m\\ \\ 0, & k > \frac{n}{2} - m. \end{cases}$$

For each $n$, the $n^{th}$ entry of the sequence we are interested in can be rewritten as

$$\sum _{k=0}^{\frac{n}{2}-m} \frac{1}{k!}(-\frac12)^k \alpha_{n,k} = \sum_{k=0}^{\infty} (S_k - S_{k-1})\alpha_{n,k} = \sum_{k=0}^{\infty} S_k (\alpha_{n,k} - \alpha_{n,k+1})$$

As long as $m > \frac14$, it is easy to check $a_{n,k}$ is a non-negative decreasing sequence in $k$ for fixed $n$.
We have:

1. For fixed $n$, $\displaystyle\quad\sum_{k=0}^{\infty} |\alpha_{n,k} - \alpha_{n,k+1}| = \alpha_{n,0} = e^{(3m-2m^2)/n} \le \max( e^{3m-2m^2}, 1 ) < \infty.$
2. $\displaystyle\lim_{n\to\infty} \sum_{k=0}^{\infty} (\alpha_{n,k} - \alpha_{n,k+1}) =\lim_{n\to\infty} a_{n,0} = \lim_{n\to\infty} e^{(3m-2m^2)/n} = 1.$

3. For fixed $k$, $\displaystyle\quad\lim_{n\to\infty} (\alpha_{n,k}-\alpha_{n,k+1}) = 1 - 1 = 0.$

The double sequence $\alpha_{n,k}-\alpha_{n,k+1}$ satisfies the conditions for Silverman-Toeplitz theorem and hence

$$\lim_{n\to\infty} \sum_{k=0}^{\infty}S_k(\alpha_{n,k}-\alpha_{n,k+1}) = \lim_{k\to\infty} S_k = \sum_{k=0}^{\infty} \frac{1}{k!}(-\frac12)^k = e^{-1/2}$$

Notes

Using Silverman-Toeplitz is not the most efficient way to prove the statement. There is a theorem

Dominated convergence theorem for series

Let $\beta_{n,k}$ be a double sequence and $b_k$, $d_k$ be two sequences such that

1. $\displaystyle \lim_{n\to\infty} \beta_{n,k} = b_k$ exists for every $k$.
2. For every $n$ and $k$, $|\beta_{n,k}| < d_k$ and $\displaystyle\;\sum_{k} d_k < \infty$.

i.e. the rows of the double sequence $\beta_{n,k}$ are dominated by a single non-negative sequence $d_k$ which has finite sum. Then

$$\sum_{k} b_k = \sum_{k} \lim_{n\to\infty} \beta_{n,k} = \lim_{n\to\infty} \sum_{k}\beta_{n,k}$$

One can verify

\begin{align} \beta_{n,k} = & \frac{1}{k!}(-\frac12)^k \alpha_{n,k}\\ b_k = & \frac{1}{k!}(-\frac12)^k\\ d_k = & \frac{1}{k! 2^k} \max( e^{3m-2m^2}, 1 ) \end{align} satisfies above condition for "DCT for series" and the statement follows immediately. I can't find any wiki page for this "DCT for series" but here is a proof I find online.

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Thanks Achille! Very neat! – Friedrich Nietzsche Sep 21 '13 at 18:33
two comments:(1) Concerning the use of DCT, we considered $m$ as a given number. Don't we have to worry about the case of $m$ depending on $n$. For example $m=n/2$, in that case, it is not true anymore that $\beta_{n,k} \rightarrow b_k$ since $\beta_{n,k} \rightarrow 0$ . Recall that $k$ goes from $0$ to $n/2 -m$. (2) In defining $d_k$, is it $\frac{1}{k! 2^k} \max( e^{2m-3m^2}, 1 )$ or $\frac{(-1/2)^k}{k!} \max( e^{2m-3m^2}, 1 )$. Thanking you! – Friedrich Nietzsche Sep 22 '13 at 15:21
1) If $m$ indeed depends on $n$, then this will become a different problem. However, that isn't the case and once $m$ is given, it only affect the definition of $\alpha_{n,k}$. 2) d_k needs to be a non-negative sequence and it is $\frac{1}{k!2^k} \max(e^{2m-3m^2},1)$. The situation is similar to that in Lebesgue dominated convergence theorem, you need a non-negative sequence/function to dominated others so that one can exchange the order of taking limit and summation/integration. – achille hui Sep 22 '13 at 15:39
I thought it would not affect the reasoning, so I shared half of the problem then. Here is my full problem $P\left[X=k\right]= \frac{{{\rm (}{{\frac{{\rm 1}}{{\rm 2}}}}{\rm )}}^k}{k{\rm !}}\sum^{\left\lfloor \frac{n}{{\rm 2}}{\rm -}k\right\rfloor }_{r{\rm =0}}{\frac{{\left({\rm -}{{\frac{{\rm 1}}{{\rm 2}}}}\right)}^r}{r{\rm !}}e^{\frac{3(r+k)-2{(r+k)}^2}{n}}}$. I want to show that $X$ converges to a Poisson (1/2). . I calculated the 8 first moments of $X$ and I found that all of them converge to those of a Poisson (1/2) hence my suspicion. – Friedrich Nietzsche Sep 22 '13 at 15:50
Find a typo in the $2^{nd}$ part of my answer, $d_k$ should be $\frac{1}{k! 2^k}\max(e^{3m-2m^2},1)$. About the problem of $X$ converges to Poisson(1/2). What we have shown is some sort of point-wise convergence. The proof can probably be extended to expectation value of any $f(X)$ where $f(x)$ falls to zero fast enough for large $x$. However, to really prove $X$ converge to Poisson(1/2), we need a proof of uniform convergence. I don't think there is any nice theorem one can directly use. – achille hui Sep 22 '13 at 16:28