TITLE: Asymptotic Expansion for $\frac1{2^n}\sum_{k=1}^n\frac1{\sqrt{k}}\binom{n}{k}$
QUESTION [14 upvotes]: Prior Art
The fact that
$$
\lim_{n\to\infty}\frac1{2^n}\sum_{k=1}^n\frac1{\sqrt{k}}\binom{n}{k}=0\tag1
$$
is the topic of this question. An argument using a bit of probability theory gives a first order estimate of the size of the sum.

Estimate of the Sum
The binomial distribution has mean $\frac n2$ and variance $\frac n4$, Chebyshev's Inequality says
$$
\frac1{2^n}\sum_{k=1}^n\binom{n}{k}\left[\left|k-\frac n2\right|\ge x\right]\le\frac n{4x^2}\tag2
$$
Setting $x=n^{7/8}$ in $(2)$ gives
$$
\frac1{2^n}\sum_{k=1}^n\binom{n}{k}\left[\left|k-\frac n2\right|\ge n^{7/8}\right]\le\frac1{4n^{3/4}}\tag3
$$
which means
$$
\frac1{2^n}\sum_{k=1}^n\binom{n}{k}\left[\left|k-\frac n2\right|\lt n^{7/8}\right]\ge1-\frac1{4n^{3/4}}\tag4
$$
Note that when $\left|k-\frac n2\right|\le n^{7/8}$,
$$
\frac1{\sqrt{n/2+n^{7/8}}}\le\frac1{\sqrt{k}}\le\frac1{\sqrt{n/2-n^{7/8}}}\tag5
$$
and the width of this interval is
$$
\begin{align}
\frac1{\sqrt{n/2-n^{7/8}}}-\frac1{\sqrt{n/2+n^{7/8}}}
&=\frac{\sqrt{n/2+n^{7/8}}-\sqrt{n/2-n^{7/8}}}{\sqrt{n^2/4-n^{7/4}}}\\
&=\frac{2n^{7/8}}{\sqrt{n^2/4-n^{7/4}}\left(\sqrt{n/2+n^{7/8}}+\sqrt{n/2-n^{7/8}}\right)}\\[6pt]
&=O\!\left(n^{-5/8}\right)\tag6
\end{align}
$$
Thus, when $\left|k-\frac n2\right|\le n^{7/8}$,
$$
\frac1{\sqrt{k}}=\sqrt{\frac2n}+O\!\left(n^{-5/8}\right)\tag7
$$
In any case, $\frac1{\sqrt{k}}\le1$.
Therefore,
$$
\begin{align}
\frac1{2^n}\sum_{k=1}^n\frac1{\sqrt{k}}\binom{n}{k}
&=\frac1{2^n}\sum_{k=1}^n\frac1{\sqrt{k}}\binom{n}{k}\left[\left|k-\frac n2\right|\le x\right]\\
&+\frac1{2^n}\sum_{k=1}^n\frac1{\sqrt{k}}\binom{n}{k}\left[\left|k-\frac n2\right|\ge x\right]\\
&=\left(\sqrt{\frac2n}+O\!\left(n^{-5/8}\right)\right)\left(1+O\!\left(n^{-3/4}\right)\right)\\[6pt]
&+O(1)\,O\!\left(n^{-3/4}\right)\\[6pt]
&=\sqrt{\frac2n}+O\!\left(n^{-5/8}\right)\tag8
\end{align}
$$

Question
Using the approach above, it is hard to see how to get more terms of an asymptotic expansion. Chebyshev is not very precise and that might be a limitation. The Binomial distribution approaches a Normal distribution quite well and that might offer a better approach. Is there an approach that allows more terms of an asymptotic expansion to be determined?

REPLY [6 votes]: The  key idea is to use the Laplace transform
$\tfrac1{\sqrt{\pi}}\int_{\mathbb{R}_+}\exp(-xt)t^{-1/2}=\tfrac1{\sqrt{x}}$
therefore, the sum in question (denoted by $S_n$) becomes, by the binomial theorem
$$
2^n S_n=(1/\sqrt{\pi}) \int_0^{\infty} t^{-1/2}\left((\exp(-t)+1)^n-1\right)\underbrace{=}_{ibp}C_n\underbrace{\int_0^{\infty}\sqrt{t}e^{-t}(1+e^{-t})^{n-1}}_{I_n}
$$
with $C_n=2 n/\sqrt{\pi}$. Now expand the $\log(1+\exp(-t))$ term around it maximum value at $t=0$ (Exercise: Justify this Laplace style argument)
$$
I_n=2^{n-1}\int_0^{\infty}\sqrt{t}e^{-(n+1)t/2}\left(1+t^2(n+1)/8+o(nt^2) \right)
$$
evaluating the Gaussians yields
$$
I_n=\sqrt{2 \pi}2^{n-1}\left(\frac{1}{n^{3/2}}+\frac{3}{8 n^{5/2}}+o(n^{-5/2})\right)
$$
or finally

$$
S_n= C_n I_n/2^n =\frac{\sqrt{2}}{n^{1/2}}\left(1+\frac{3}{8 n}+o(n^{-1})\right)
$$

of course this approch easily generalizes to higher orders ….