<s>[INST] Limit of a sequence defined by a sum: $\lim_{n\to \infty} \frac1{2^n}\sum_{k=1}^n \frac1{\sqrt k}\binom nk$ $\lim_{n\to \infty} \frac{1}{2^n}\sum_{k=1}^n \frac{1}{\sqrt k}\binom{n}{k}$
Could it have a connection to Euler summability?                        
 [/INST] $$
\frac{1}{2^n}\sum_{k=1}^n \frac{1}{\sqrt k}\binom{n}{k} = \frac{1}{2^n}\sum_{k=1}^{\log n-1} \frac{1}{\sqrt k}\binom{n}{k} + \frac{1}{2^n}\sum_{k=\log n}^{n} \frac{1}{\sqrt k}\binom{n}{k}
$$
The second term is upper bounded by 
$$
\frac{1}{\sqrt{\log n}}\cdot\frac{1}{2^n}\sum_{k=\log n}^{n} \binom{n}{k} \leq \frac{1}{\sqrt{\log n}}\cdot\frac{1}{2^n}\sum_{k=0}^{n} \binom{n}{k} = \frac{1}{\sqrt{\log n}}\cdot 1 \xrightarrow[n\to\infty]{} 0
$$ 
and the first one by
$$
\frac{1}{2^n}\sum_{k=1}^{\log n-1} \binom{n}{k} \leq \log n\cdot\frac{1}{2^n} \binom{n}{\log n}
$$ 
What can you say about this last thing?
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