Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is it true that:

$$ \lim_{n \to \infty} \sum_{k=1}^{n-1}\binom{n}{k}2^{-k(n-k)} = 0 \;?$$

It seems true numerically, but how can this limit be shown?

share|improve this question

1 Answer 1

up vote 2 down vote accepted

Note that $(n-k)$ is at least $n/2$ for $k$ between 1 and $n/2$. Then, looking at the sum up to $n/2$ and doubling bounds what you have above by something like:

$$\sum_{k=1}^{n-1}\binom{n}{k}2^{-kn/2}=\left(1+2^{-n/2}\right)^n-2^{-n/2}-1$$

which bounds your sum above and goes to zero.

Alternatively, use the bound

$$\binom{n}{k}\leq \frac{n^k}{k!}\;.$$

Since the sum is symmetric around $k=n/2$, work with the sum up to $n/2$. Then $n^k2^{-k(n-k)}=2^{k(\log_2 n-n+k)}$. For $k$ between 1 and $n/2$ and for large $n$ this scales something like $2^{-kn/2}$, which when summed from 1 to $n/2$ in $k$ will tend to 0 as $n\rightarrow\infty$.

share|improve this answer
    
@BrianM.Scott: Whoops! Let me try to correct. –  Alex R. Dec 3 '12 at 2:57

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.