Limit of $ u_{n}=\sin(\frac{1}{n+1})+\cdots+\sin(\frac{1}{2n})$ How can I find the limit of 
$$ u_n =\sin\left(\frac{1}{n+1}\right)+\cdots+\sin\left(\frac{1}{2n}\right)$$
when $n\rightarrow\infty$?
We have:
$$ \sum_{n=1}^\infty u_{n+1}-u_n =u_\infty -\sin\left(\frac{1}{2}\right)$$
So how can I find $$ \sum_{n=1}^\infty u_{n+1}-u_n =\sum_{n=1}^\infty  \sin\left(\frac{1}{2n+2}\right)+\sin\left(\frac{1}{2n+1}\right)-\sin\left(\frac{1}{n+1}\right)\  ?$$

We can make use of the fact that $$\sin x = \sum\limits_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{(2n+1)!}$$
and note that when $x\leq 1$, the absolute value of each term is more than the sum of the absolute values of the later terms. Thus $x-\frac{x^3}{6}<\sin x<x$ for $x\leq 1$, so we have $$\sum\limits_{k=n+1}^{2n}\frac{1}{k}-n\frac{1}{6n^3}<\sum\limits_{k=n+1}^{2n}\frac{1}{k}-\frac{1}{6k^3}<u_n<\sum\limits_{k=n+1}^{2n}\frac{1}{k}$$
and $$\begin{eqnarray}\lim\limits_{n\to\infty}\sum\limits_{k=n+1}^{2n}\frac{1}{k}&=&\lim\limits_{n\to\infty}\left(\sum\limits_{k=1}^{2n}\frac{1}{k}-\ln(2n)\right)-\lim\limits_{n\to\infty}\left(\sum\limits_{k=1}^{n}\frac{1}{k}-\ln(n)\right)+\lim\limits_{n\to\infty}(\ln(2n)-\ln(n))\\
&=&\lim\limits_{n\to\infty}(\ln(2n)-\ln(n))\\
&=&\lim\limits_{n\to\infty}\ln 2\\
&=&\ln 2\end{eqnarray}$$
while clearly $\lim\limits_{n\to\infty}n\frac{1}{6n^3}=0$, so by squeeze theorem $\lim\limits_{n\to\infty} u_n=\ln 2$.
