TITLE: Limit of $\sum_{i=1}^n \left(\frac{{n \choose i}}{2^{in}}\sum_{j=0}^i {i \choose j}^{n+1}\right)$
QUESTION [17 upvotes]: I'm trying to calculate the limit for the sum of binomial coefficients:
$$S_{n}=\sum_{i=1}^n \left(\frac{{n \choose i}}{2^{in}}\sum_{j=0}^i {i \choose j}^{n+1} \right).$$

REPLY [5 votes]: Simply multiplying the largest term by the number of terms in the sum yields
$$
\begin{align}
\frac{\binom{n}{i}}{2^{in}}\sum_{j=0}^i\binom{i}{j}^{n+1}
&\le(i+1)\binom{i}{i/2}\binom{n}{i}\left(\binom{i}{i/2}2^{-i}\right)^n\tag{1}
\end{align}
$$
using $\binom{i}{i/2}=\binom{i}{(i\pm1)/2}=\frac12\binom{i+1}{(i+1)/2}$ for odd $i$.
Since $\sum\limits_{j=0}^i\binom{i}{j}=2^i$, we know that
$$
\binom{i}{i/2}\le2^i\tag{2}
$$
Furthermore, $\binom{i}{i/2}2^{-i}$ is non-increasing:
$$
\begin{align}
\binom{2k-1}{k-1}2^{-2k+1}
&=\binom{2k}{k}2^{-2k}\\[6pt]
&=\frac{2k+2}{2k+1}\binom{2k+1}{k+1}2^{-2k-1}\\[6pt]
&\gt\binom{2k+1}{k+1}2^{-2k-1}\tag{3}
\end{align}
$$
Note that
$$
\sum\limits_{i=0}^n(i+1)\ 2^i\,\binom{n}{i}=\frac{2n+3}{3}3^n\tag{4}
$$
Combining $(1)$, $(2)$, $(3)$, and $(4)$ yields
$$
\begin{align}
\sum_{i=k}^n\frac{\binom{n}{i}}{2^{in}}\sum_{j=0}^i\binom{i}{j}^{n+1}
&\le\sum_{i=k}^n(i+1)\binom{i}{i/2}\binom{n}{i}\left(\binom{i}{i/2}2^{-i}\right)^n\\
&\le\sum_{i=k}^n(i+1)\ \quad2^i\quad\binom{n}{i}\left(\binom{i}{i/2}2^{-i}\right)^n\\
&\le\frac{2n+3}{3}3^n\left(\binom{k}{k/2}2^{-k}\right)^n\tag{5}
\end{align}
$$
Applying $(5)$ with $k=23$ gives
$$
\begin{align}
\sum_{i=23}^n\frac{\binom{n}{i}}{2^{in}}\sum_{j=0}^i\binom{i}{j}^{n+1}
&\le\frac{2n+3}{3}0.48354^n\\
&=o(2^{-n})\tag{6}
\end{align}
$$
For $i=1$,
$$
\frac{\binom{n}{i}}{2^{in}}\sum\limits_{j=0}^i\binom{i}{j}^{n+1}=\frac{2n}{2^n}\tag{7}
$$
For $i=2$,
$$
\frac{\binom{n}{i}}{2^{in}}\sum\limits_{j=0}^i\binom{i}{j}^{n+1}=\frac{n(n-1)/2}{4^n}\left(2+2^{n+1}\right)\sim\frac{n^2-n}{2^n}\tag{8}
$$
For $i\ge3$,
$$
\begin{align}
\frac{\binom{n}{i}}{2^{in}}\sum\limits_{j=0}^i\binom{i}{j}^{n+1}
&\le(i+1)\binom{i}{i/2}\binom{n}{i}\left(\binom{i}{i/2}2^{-i}\right)^n\\
&=O\left(n^4\left(\frac38\right)^n\right)\\[6pt]
&=o(2^{-n})\tag{9}
\end{align}
$$
because $\binom{n}{i}\le\frac{n^i}{i!}$ and for $i\ge5$, $\binom{i}{i/2}2^{-i}\lt\frac38$.
Thus, $(6)$, $(7)$, $(8)$, and $(9)$ show that
$$
S_n\sim\frac{n^2+n}{2^n}\tag{10}
$$