Bound for $\log { \binom{n}{i}}$?

Question

(1) Are there better (smaller; tighter) bounds for $\log { \binom{n}{i}}$, than $O(n \log n )$?

(2) Under what conditions $O(i \log n)$ is a good bound? Clearly this bound should be in a way that it respects symmetricity of this quantity, i.e. treat $i = j$ and $i = n-j$ the same, for any $j \in \{0, 1, 2, ..., n\} $.

(3) What causes the discrepancy between the following two bounds for $A = \sum_{i=1}^M \log (1+m/i)$?:

3.i
$$ \sum_{i=1}^M \log (1+m/i) = \log \frac{(M+m)!}{M! \times m!} \in O((M+m) \log(M+m) ) $$

3.ii

Since $\sum_{i=1}^p 1/i \approx \log p$, we can write: $$ \sum_{j=i}^{i+m} 1/j \approx \log \frac{m+i}{i} = \log (1+m/i) $$ Hence: $$ A = \sum_{i=1}^M \log (1+m/i) = \sum_{i=1}^M \sum_{j=i}^{i+m} 1/j = m \sum_{i=1}^M 1/i = m \log M $$

Answer 1

$n\log n$ is not a very good bound for $\log \binom {n}{i}.$ Use Stirling's formula : $n!=[1+d(n)] (n/e)^n \sqrt {2 \pi n}$ where $|d(n)|<1/4 n$ for $n>0$. The largest value of $\binom {n}{i}$, which is $\binom {n}{n/2}$ for even $n$, and $\binom {n}{(n-1)/2}$ for odd $n$, is $[1+f(n)]2^n/\sqrt {\pi n}$ where $f(n)\to 0$ as $n\to \infty$.

Answer 2

Easier: $\binom{n}{k} \sim \frac{n^k}{k!}$, so the exptession is $O(\log n)$