How to calculate the limit that seems very complex.. Someone gives me a limit about trigonometric function and combinatorial numbers.
$I=\displaystyle \lim_{n\to\infty}\left(\frac{\sin\frac{1}{n^2}+\binom{n}{1}\sin\frac{2}{n^2}+\binom{n}{2}\sin\frac{3}{n^2}\cdots\binom{n}{n}\sin\frac{n+1}{n^2}}{\cos\frac{1}{n^2}+\binom{n}{1}\cos\frac{2}{n^2}+\binom{n}{2}\cos\frac{3}{n^2}\cdots\binom{n}{n}\cos\frac{n+1}{n^2}}+1\right)^n$
when $n$ is big enough, $\displaystyle 0\leqslant\frac{n+1}{n^2}\leqslant \frac\pi 2$ I tried $\displaystyle \frac{x}{1+x}\leqslant\sin x\leqslant x$ 
Use $\sin x\leqslant x$, I got $I\leqslant e$, But use another inequality I got nothing. 
I don't know the answer is $e$ or not. Who can help me. Thanks.

Consider $$\lim_{n\to\infty} n\frac{\displaystyle\sum\limits_{k=0}^n {n \choose k}\sin\frac{k+1}{n^2}}{\displaystyle\sum\limits_{k=0}^n {n \choose k}\cos\frac{k+1}{n^2}}.$$ If this limit equals $\alpha$, your limit equals $e^\alpha$. In order to simplify a bit, thanks to the Taylor expansions of $\sin$ and $\cos$ near $0$,  we may multiply top and bottom by $n^2$: $$\begin{align} \lim_{n\to\infty} n\frac{\displaystyle\sum\limits_{k=0}^n {n \choose k}n^2\sin\frac{k+1}{n^2}}{\displaystyle\sum\limits_{k=0}^n {n \choose k}n^2\cos\frac{k+1}{n^2}}&=\lim_{n\to\infty} {n}\frac{\displaystyle\sum\limits_{k=0}^n {n \choose k}(k+1)}{\displaystyle\sum\limits_{k=0}^n {n \choose k}\left(n^2-\frac{1}{2}\frac{(k+1)^2}{n^2}\right)} \\ &= \lim_{n\to\infty} {n}\frac{\displaystyle\sum\limits_{k=0}^n k{n \choose k} + \displaystyle\sum\limits_{k=0}^n {n \choose k}}{n^2\displaystyle\sum\limits_{k=0}^n {n \choose k} - \frac{1}{2n^2}\displaystyle\sum\limits_{k=0}^n (k+1)^2{n \choose k}} \\ &= \lim_{n\to\infty} \frac{2^{n-1}n(n+2)}{n^2 2^n-2^{n-3}(1+5/n+4/n^2)}=\frac{1}{2}.\end{align}$$ Thus your expression approaches $\sqrt{e}$.
