Determine if $\sum^{\infty}_{n=1}\int^{\frac{\sin n}{n}}_0\frac{\sin x}{x} \, dx $ is converges or diverges. Determine if $\sum^{\infty}_{n=1}\int^{\frac{\sin n}{n}}_0\frac{\sin x}{x} \, dx $ is converges or diverges. 
Could someone please show me how to do it? And especialy showing why $\lim_{n\to\infty}\int^{\frac{\sin n}{n}}_0\frac{\sin x}{x} \, dx=0$ 
(I know that $\lim_{n\to\infty} \frac{\sin n}{n}=0$)
 A: This is edited version. Thanks to Tigran Saluev for pointing out the mistatke.
By his argument your series converges iff the series 
$$
\sum\limits_{n=1}^\infty\frac{|\sin n|}{n}\tag{1}
$$
converges.
Note that 
$$
\sum\limits_{n=1}^\infty\frac{|\sin n|}{n}\geq\sum\limits_{n=1}^\infty\frac{\sin^2 n}{n}=\frac{1}{2}\sum\limits_{n=1}^\infty\frac{1}{n}-\frac{1}{2}\sum\limits_{n=1}^\infty\frac{\cos 2n}{n}
$$
The first term diverges as harmonic series diverges. The second term converges since
$$
\sum\limits_{n=1}^\infty\frac{\cos 2n}{n}
=\Re\left(\sum\limits_{n=1}^\infty\frac{e^{i\cdot 2n}}{n}\right)
=\Re\left(\sum\limits_{n=1}^\infty\frac{z^n}{n}\biggl|_{z=e^{2i}}\right)
=\Re\left(-\ln(1-z)|_{z=e^{2i}}\right)<\infty
$$
Thus the series $(1)$ diverges, hence your series diverges too.
A: $\frac{\sin x}{x}$ is a continuous function, thus
$$\lim_{a\rightarrow 0}\int_0^a \frac{\sin x}{x}\,dx\ \ \ \mbox{ is always equal to } 0.$$
But that is not enough.
$\frac{\sin x}{x}$ is also positive in a neighborhood of 0, thus all the integrals for $n$ greater than some fixed $N$ are positive. It doesn't depend on the sign of $\frac{\sin n}{n}$, since the integrated function is even. But the fact $\frac{\sin n}{n}$ is not necessarily positive forbids you to write 
$$\int_0^{\frac{\sin n}{n}} \frac{\sin x}{x}\,dx \le \frac{\sin n}{n}
$$
like Norbert has done. Correct estimate looks like
$$\int_0^{\frac{\sin n}{n}} \frac{\sin x}{x}\,dx \le \frac{|\sin n|}{n}.$$
So the solution is not that simple.
$\frac{\sin x}{x}$ is decreasing and convex on $[0, 1]$:

So we can estimate the integral from below as the square of highlighted region:
$$
\int_0^{\frac{\sin n}{n}} \frac{\sin x}{x}\,dx \ge \left.\frac{\sin x}{x}\right|_{x=\frac{|\sin n|}{n}}\cdot\frac{|\sin n|}{n} + \frac{1}{2}\cdot\frac{|\sin n|}{n}\cdot\left(1 - \left.\frac{\sin x}{x}\right|_{x = \frac{\sin n}{n}}\right) =
$$
$$
= \frac{1}{2}\sin\left(\frac{|\sin n|}{n}\right) + \frac{1}{2}\frac{|\sin n|}{n}.
$$
Thus, including my remark on Norbert's answer, your series converges if and only if $$\sum_{n=1}^{\infty}\frac{|\sin n|}{n}$$ converges.
A: Hint for the second part of your question: The mean-value theorem shows that  $\int_0^{\frac{\sin{n}}{n}} \frac{\sin{x}}{x}~dx=\frac{\sin{y_n}}{y_n}\frac{\sin{n}}{n}$ for some $0\leq y_n\leq\frac{\sin{n}}{n}$. What can you deduce? What du you know about the behaviour of $\frac{\sin{x}}{x}$ near $0$?
A: Let $I_n=\int_{0}^{\frac{\sin n}{n}}{\frac{\sin x}{x}}dx$.
$I_n$ is between $a_n=\int_{0}^{\frac{\sin n}{n}}{\frac{x}{x}}dx=\frac{\sin n}{n}$ and $\int_{0}^{\frac{\sin n}{n}}{\frac{x-\frac{x^3}{6}}{x}}dx=\frac{\sin n}{n}-\frac{\left(\sin n\right)^3}{18n^3}$, and the difference $|I_n-a_n|\le\frac{\left|\sin n\right|^3}{18n^3}\le \frac{1}{n^3}$, whose sum converges absolutely.
So $\sum_{0}^{\infty}{I_n}$ converges (absolutely) if and only if $\sum_{0}^{\infty}{a_n}=\sum_{0}^{\infty}{\frac{\sin n}{n}}$ converges (absolutely).
And $\sum_{0}^{\infty}{\frac{\sin n}{n}}$ converges (Does $\sum\frac{\sin n}{n}$ converge?), but not absolutely.
