Asymptotic Expansion for $\frac1{2^n}\sum_{k=1}^n\frac1{\sqrt{k}}\binom{n}{k}$

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.

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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, $$ \begin{align} \frac1{2^n}\sum_{k=1}^n\frac1{\sqrt{k}}\binom{n}{k} &=\frac1{2^n}\sum_{k=1}^n\frac1{\sqrt{k}}\binom{n}{k}\left[\left|k-\frac n2\right|\le x\right]\\ &+\frac1{2^n}\sum_{k=1}^n\frac1{\sqrt{k}}\binom{n}{k}\left[\left|k-\frac n2\right|\ge x\right]\\ &=\left(\sqrt{\frac2n}+O\!\left(n^{-5/8}\right)\right)\left(1+O\!\left(n^{-3/4}\right)\right)\\[6pt] &+O(1)\,O\!\left(n^{-3/4}\right)\\[6pt] &=\sqrt{\frac2n}+O\!\left(n^{-5/8}\right)\tag8 \end{align} $$

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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?

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