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 $$ \sum_{n=1}^\infty u_{n+1}-u_n =\sum_{n=1}^\infty  \sin\left(\frac{1}{2n+2}\right)+\sin\left(\frac{1}{2n+1}\right)-\sin\left(\frac{1}{n+1}\right)\  ?$$
 A: Using
$$ x - \frac{x^3}{6} \lt \sin x \lt x$$
for positive $x$ close to $0$ and the fact that $\displaystyle \sum_{k=1}^{n} \frac{1}{k^3}$ is convergent, we see that
your limit is the same as the limit of $\displaystyle s_n = \sum_{k=1}^{n} \frac{1}{n+k}$
One method to find this limit is to use the Harmonic series estimate as in Alex's answer.
Another method is to interpret it as a Riemann sum of the integral $\int_{0}^{1} \frac{1}{1+x} \text{ dx}$ as in Ragib's comment to Alex's answer.
I will mention a third method, which uses the Mean Value theorem. (though all three are quite similar).
Since $\frac{1}{x}$ is decreasing for positive $x$, we see using that mean value theorem that
$$\frac{1}{x+1} \lt \log (1+x) - \log x \lt \frac{1}{x}$$
Setting $x = n+1, n+2, \dots, 2n$ and adding, and then setting $x=2n-1, 2n-2, \dots, n$ and adding we get
$$\log \left(\frac{2n+1}{n}\right) \lt s_n \lt \log \left(\frac{2n-1}{n-1}\right)$$
and thus 
$$\lim_{n \to \infty} s_n = \log 2$$
A: 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 $\lim\limits_{n\to\infty} u_n=\ln 2$.
