Does $\sum\int\arctan(n^2x)\sin(1/x)\,dx$ converge? Does the series
$$\sum_{n = 1}^\infty\int_0^{\frac{1}{n}} \arctan(n^2 x)\sin\left(\frac{1}{x}\right)dx$$
converge?
I have tried to estimate this integral, but I can't get upper estimate better than $\frac{1}{n}$. Can you give me a hint, please?
 A: We will show that the summand 
$$
a_{n} = \int_{0}^{1/n} \arctan(n^{2}x)\sin \left(\frac{1}{x}\right)dx = \int_n^{\infty} \arctan\left(\frac{n^{2}}{y}\right) \frac{\sin(y)}{y^{2}}dy
$$
satisfies $|a_{n}|\leq C/n^{2}$ for some $C>0$, so the series converges absolutely. The key idea is that the integral is oscilating, so it cancels out a lot. We need to control it. 
Choose $N$ such that $2N\pi \geq n$, so that $N = \lceil{n/2\pi\rceil}$. 
Then we can write
$$
a_{n} = \epsilon_{n} + \sum_{k=N}^{\infty} b_{k}
$$
where 
$$
\epsilon_{n} = \int_{n}^{2\pi N}\arctan\left(\frac{n^{2}}{y}\right) \frac{\sin(y)}{y^{2}} dy \\
b_{k} = \int_{2k\pi}^{(2k+2)\pi}\arctan\left(\frac{n^{2}}{y}\right)\frac{\sin(y)}{y}
$$
From $2\pi N \leq n + 2\pi$, it is easy to check that $|\epsilon_{n}| \leq C_{1}/n^{2}$ for some $C_{1}>0$.  For $b_{k}$, by splitting the integration interval as $[2k\pi, (2k+1)\pi]$ and $[(2k+1)\pi, (2k+2)\pi]$ and using $\sin(y+\pi) = -\sin(y)$, we get
$$
b_{k} = \int_{2k\pi}^{(2k+1)\pi}(f(y) - f(y+\pi)) \sin(y) dy
$$
where
$$
f(y) = \frac{1}{y^{2}} \arctan\left(\frac{n^{2}}{y}\right).
$$
By the mean value theorem, we can bound $|f(y)-f(y+\pi)|$ and we get
$$
|b_{k}| \leq \frac{1}{16k^{3}} + \frac{1}{4n^{4}k^{2}}
$$
which proves
$$
|a_{n}| \leq O\left(\frac{1}{n^{2}}\right) + \sum_{k=N}^{\infty} \left(\frac{1}{16k^{3}} + \frac{1}{4n^{4}k^{2}}\right)
\\
= O\left(\frac{1}{n^{2}}\right)
$$
