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Can someone please help me simplifying this sum

$\sum_{k=0}^n \binom{n}{k} 2^{k^2}$

Wolframalpha fails (see here).

Thanks in advance.

The sum counts the number of (labelled) digraphs (with loops) with vertex set $V \subseteq [n]$.

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    I think you entered it wrong, Wolframalpha seems to read it as the bracket $\{ k=0\}$ is missing. Then it becomes :$ \sum_k=0^n \binom{n}{k} 2^{k^2}$.2011-10-25

1 Answers 1

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I agree with J.M.'s comment above that this is not going to resolve into something simple. Asymptotically, though, the last term will dominate, and maybe that will be useful. We have $2^{n^2} \leq \sum_{k=0}^n \binom{n}{k} 2^{k^2} \leq 2^{n^2}\left(1 + \frac{2n^2}{4^n}\right).$ The relative remainder term $R(n) = \frac{2n^2}{4^n}$ goes to $0$ fairly quickly as $n \to \infty$, and so $\sum_{k=0}^n \binom{n}{k} 2^{k^2} \approx 2^{n^2}$.

Derivation: The terms in the sum are strictly increasing. Since $n \geq k$, for $k \geq 1$ we have $\frac{\binom{n}{k} 2^{k^2}}{\binom{n}{k-1}2^{(k-1)^2}} = 2^{2k-1}\left(\frac{n}{k}-1+\frac{1}{k}\right) \geq 2^{2k-1}\left(1-1+\frac{1}{k}\right) = \frac{2^{2k}}{2k}> 1.$

Therefore, $2^{n^2} \leq \sum_{k=0}^n \binom{n}{k} 2^{k^2} = 2^{n^2} + \sum_{k=0}^{n-1} \binom{n}{k} 2^{k^2} \leq 2^{n^2} + \sum_{k=0}^{n-1} \binom{n}{n-1} 2^{(n-1)^2} = 2^{n^2} + n^2 2^{n^2-2n+1}$ $= 2^{n^2}\left(1 + \frac{2n^2}{4^n}\right).$

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    @Hans: You're welcome! With respect to your question, I don't know. And while I don't think there is a nice closed-form answer, I can't prove that there isn't one, either.2011-10-26