Limit of a sequence defined by a sum: $\lim_{n\to \infty} \frac1{2^n}\sum_{k=1}^n \frac1{\sqrt k}\binom nk$

Question

$\lim_{n\to \infty} \frac{1}{2^n}\sum_{k=1}^n \frac{1}{\sqrt k}\binom{n}{k}$

Could it have a connection to Euler summability?                        

Answer 1

Here is another approach using Cauchy-Schwarz: We have \begin{align} \sum_{k=1}^n \dfrac1{\sqrt{k}} \cdot \dbinom{n}k & \leq \sqrt{\sum_{k=1}^n \dfrac1{\sqrt{k}^2}} \cdot \sqrt{\sum_{k=1}^n \dbinom{n}k^2}\\ & = \sqrt{\sum_{k=1}^n \dfrac1k} \cdot \sqrt{\dbinom{2n}n-1}\\ & \leq \sqrt{\ln(n)+1} \cdot \sqrt{\dbinom{2n}n} \sim \sqrt{\ln(n)} \sqrt{\dfrac{4^n}{\sqrt{\pi n}}} \end{align} Hence, $$\dfrac1{2^n}\sum_{k=1}^n \dfrac1{\sqrt{k}} \cdot \dbinom{n}k \in \mathcal{O}\left( \sqrt{\dfrac{\ln(n)}{\sqrt{\pi n}}}\right)$$ Hence, $$\lim_{n \to \infty} \dfrac1{2^n}\sum_{k=1}^n \dfrac1{\sqrt{k}} \cdot \dbinom{n}k = 0$$

Answer 2

$$ \frac{1}{2^n}\sum_{k=1}^n \frac{1}{\sqrt k}\binom{n}{k} = \frac{1}{2^n}\sum_{k=1}^{\log n-1} \frac{1}{\sqrt k}\binom{n}{k} + \frac{1}{2^n}\sum_{k=\log n}^{n} \frac{1}{\sqrt k}\binom{n}{k} $$ 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?

Answer 3

Another approach, more self-contained (no need to know anything about Stirling, or non-trivial properties of the binomial coefficients)

Fix any $\epsilon > 0$, and let $N_\epsilon\geq 1$ be an integer such that $\frac{1}{\sqrt{N_\epsilon}} < \epsilon$. Then, for any $n\geq N_\epsilon$, we can write $$ \frac{1}{2^n}\sum_{k=1}^n \frac{1}{\sqrt k}\binom{n}{k} = \underbrace{\frac{1}{2^n}\sum_{k=1}^{N_\epsilon-1} \frac{1}{\sqrt k}\binom{n}{k}}_A + \underbrace{\frac{1}{2^n}\sum_{k=N_\epsilon}^{n} \frac{1}{\sqrt k}\binom{n}{k}}_B $$ Now, it is easy to see that $A\xrightarrow[n\to\infty]{} 0$, as there are constantly many terms (i.e., $N_\epsilon - 1$), each of them going to $0$ (recall that for any $0\leq k \leq N_\epsilon$, $\frac{1}{\sqrt k}\binom{n}{k}\leq \binom{n}{k} \leq n^k$, and you divide by $2^n$). In other terms, there exists $N^\ast\geq 1$ such that for all $n \geq N^\ast$ one has $A \leq \epsilon$.

It remains to bound $B$: by choice of $N_\epsilon$, $$ B \leq \epsilon\cdot\frac{1}{2^n}\sum_{k=N_\epsilon}^{n}\binom{n}{k} \leq \epsilon\cdot\frac{1}{2^n}\sum_{k=1}^{n}\binom{n}{k} = \epsilon $$

Putting it together, for any $\epsilon > 0$ there exists $N=\max(N_\epsilon, N^\ast)$ such that for all $n\geq N$ $$ \frac{1}{2^n}\sum_{k=1}^n \frac{1}{\sqrt k}\binom{n}{k} \leq A+B \leq 2\epsilon $$ proving that $\frac{1}{2^n}\sum_{k=1}^n \frac{1}{\sqrt k}\binom{n}{k} \xrightarrow[n\to\infty]{} 0$.