Binomial coefficients summation I have the following expression,
$$
\frac{1}{n 4^n} \sum_{i = 1}^{n} i \binom{2n}{n + i},
$$
and I'm wondering if one can get either an exact formula for the sum involving binomial coefficient or get a tight asymptotic bound for the whole expression.
Any help would be appreciated.
 A: I have worked on a similar expression:
$$
\begin{align}
\sum_{j=1}^{n}{2j\cdot\binom{2n}{n+j}}&=\sum_{j=1}^{n}{(n+j)\cdot\binom{2n}{n+j}}-\sum_{j=1}^{n}{(n-j)\cdot\binom{2n}{n+j}}\\
\\
&=\sum_{j=1}^{n}{(n+j)\cdot\binom{2n}{n+j}}-\sum_{j=1}^{n}{(n-j)\cdot\binom{2n}{n-j}}
\end{align}
$$
Imagine choosing several of $2n$ people to be in a committee and then electing one of the committee as a leader. The first term on right hand side is the number of possibilities where the committees consist of at least $n+1$ members while the other term is where the committees consist of at most $n-1$ members.
We can select the leader first and then complete the committees. If there are at least $n+1$ members of committee, then there are at most $n-1$ non committee. If there are at most $n-1$ members, then there are at most $n-2$ members aside from the leader.
$$
\begin{align}
\sum_{j=1}^{n}{2j\cdot\binom{2n}{n+j}}&=2n\sum_{k=0}^{n-1}{\binom{2n-1}{k}}-2n\sum_{k=0}^{n-2}{\binom{2n-1}{k}}\\
\\
&=2n\cdot\binom{2n-1}{n-1}
\end{align}
$$
From this result we can get solution to your problem:
$$
\begin{align}
\frac{1}{n\cdot 4^{n}}\sum_{j=1}^{n}{j\cdot\binom{2n}{n+j}}&=\frac{1}{2n\cdot 4^{n}}\sum_{j=1}^{n}{2j\cdot\binom{2n}{n+j}}\\
\\
&=\frac{1}{2n\cdot 4^{n}}\cdot 2n\cdot\binom{2n-1}{n-1}\\
\\
&=\frac{1}{4^{n}}\cdot\binom{2n-1}{n-1}
\end{align}
$$
A: If you want approximations or bounds, use the result given by Wolfram Alpha.
$$S_n=\frac{1}{n\, 4^n} \sum_{i = 1}^{n} i \binom{2n}{n + i}=\frac{4^{-n} n \Gamma (2 n)}{\Gamma (n+1)^2}$$ Take logarithms and use Stirling approximation to obtain
$$\log\left(S_n\right)=-\frac{1}{2} \log (4 \pi  n)-\frac{1}{8 n}+\frac{1}{192 n^3}+O\left(\frac{1}{n^5}\right)$$ Continue with Taylor using
$$S_n=e^{\log\left(S_n\right)}=\frac{1}{2 \sqrt{\pi n} }\Bigg[1-\frac{1}{8 n}+\frac{1}{128 n^2}+O\left(\frac{1}{n^3}\right)\Bigg]$$
