Convergence of series of definite integrals I encountered this terrible series which I'm supposed to check for convergence.
$$ \sum_{n=1}^\infty \int_0^\frac{\sin n}{n}\frac{\sin x}{x}\,\mathrm dx $$
I can see the necessary condition $a_n \to 0$ applies, but that's not enough. Integrating this thing doesn't give pretty results, and seeing as this is supposed to be a simple exercise in convergence tests, I feel like I'm missing something. None of the common tests (ratio, root, d'Alembert, Abel, etc.) seem to give an answer to this.
Any help?
 A: Note that
$$1-\frac{\sin x}{x}=\int_0^1(1-\cos(xt))\,dt=2\int_0^1\sin^2(xt/2)\,dt
$$
Thus
$$0\le 1-\frac{\sin x}{x}\le\frac{x^2}{2}\int_0^1t^2\,dt=\frac{x^2}{6}
$$
Integrating we get
$$\left\vert \frac{\sin n}{n}-\int_0^{\sin n/n}\frac{\sin x}{x}dx\right\vert \le\frac{\sin^3 n}{24n^3}\le \frac{1}{24n^3}$$
So, the series $\sum a_n$ with
$$a_n= \frac{\sin n}{n}-\int_0^{\sin n/n}\frac{\sin x}{x}dx$$
is absolutely convergent, because $a_n=O(1/n^3)$.
On the other hand the series $\sum b_n$ with
$$b_n= \frac{\sin n}{n}$$
is semi-convergent using Abel's test. It follows that the series $\sum(b_n-a_n)$ is also convergent. But this is the series under consideration.
A: Denote $$\text{Si}(x) = \int_0^x \frac{\sin t}{t} dt$$
You can show that, using the Taylor series of $\sin t$ at the origin, $$\text{Si}(x) = x + O(x^3) $$as $x\to 0$.
Thus $$\sum_{n=1}^{\infty} \text{Si}\left(\frac{\sin n}{n}\right) = \sum_{n=1}^{\infty} \left[\frac{\sin n}{n} + O(\frac{\sin^3 n}{n^3}) \right]$$
Since $\sum \sin n / n$ is conditionally convergent, and the big-$O$ term is absolutely convergent, the desired series converges conditionally.
