TITLE: Asymptotics for sum of binomial coefficients from Concrete Mathematics
QUESTION [7 upvotes]: Concrete Mathematics EXERCISE 9.25:

Supposing
  \[ S_n = \sum_{k=0}^n \binom{3n}k \]
  Prove that
  \[ S_n = \binom{3n}{n}\left(2-\frac4n+O\left(\frac1{n^2}\right)\right) \]

This sequence also appears in OEIS A066380
I have been trying to understand the answer to the problem, but failed:

\[S_n\left/\binom{3n}n\right. = \sum_{k=0}^n \frac{n\cdots(n-k+1)}{(2n+1)\cdots(2n+k)}\tag1\]
  We may restrict the range of summation to $0 \le k \le (\log n)^2$, say. In this range $n\cdots(n-k+1) = n^k\left(1-\binom k2/n+O(k^4/n^2)\right)$ and $(2n+1)\cdots(2n+k) = (2n)^k\left(1+\binom{k+1}2/2n+O(k^4/n^2)\right)$, so the summand is
  \[ \frac1{2^k}\left(1-\frac{3k^2-k}{4n}+O\left(\frac{k^4}{n^2}\right)\right) \tag2 \]
  Hence the sum over $k$ is $2-4/n+O(1/n^2)\tag3$ Q.E.D.

The formula (1) is acceptable, because
\[ \left. \binom{3n}{n-k} \right/ \binom{3n}{n} = \frac{n\cdots(n-k+1)}{(2n+1)\cdots(2n+k)} \]
The equation (2) maybe holds for $0 \le k \le (\log n)^2$, but formula (3) seems too strange (notice that $k$ is restricted, not over integers from $[0..n]$. How can we conclude that?
I have tried to considered equation (2) as the partial sum of a power series (the Taylor series for $n^{-1}$), but there seems no evidence that the corresponding power series of (2) or (3) converges.
Now OP has understood the answer. A trivial trick is necessary. OP will look for someone clever to give a complete solution and set his/her answer as accepted answer.

REPLY [2 votes]: The claim is that $$\sum_{k=0}^{m} \frac1{2^k}\left(1-\frac{3k^2-k}{4n}+O\left(\frac{k^4}{n^2}\right)\right) = 2-4/n+O(1/n^2) $$ where $m=\lfloor \log_2^2 n \rfloor.$
Computing one term at a time: $ \displaystyle A(m)= \sum_{k=0}^m 1/2^k = 2 - 2^{-m}= 2- \frac{1}{n^{\log n}}= 2 + \mathcal{O}(n^{-2}).$
This far into the book you should know how to compute $\displaystyle \sum_{k=0}^m \frac{3k^2-k}{2^k} = \frac{ 2^{m+4} -3m^2-11m-16}{2^m}.$ (In case you forgot, try differentiating $\sum x^m/2^m$.) The only thing that survives the $\mathcal{O}(n^{-2})$ war is $2^4=16$ so the second term contributes $-4/n + \mathcal{O}(n^{-2}).$
And finally, $\displaystyle \sum_{k=1}^{\infty} \frac{k^4}{2^k}$ is convergent so the last terms contribution is certainly $\mathcal{O}(n^{-2}).$ Hence the result.