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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?

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    $\begingroup$ Where did you get this from? That might help in figuring out what tools are expected to evaluate this. $\endgroup$ – Don Thousand May 1 at 18:25
  • $\begingroup$ It is the task to prepare for the test, I don't know where it's from, but I'm quite sure that we must estimate this intergral, because Dirichlet's and Abel's test don't help $\endgroup$ – ErlGrey May 1 at 19:14
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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 $$ |a_{n}| \leq O\left(\frac{1}{n^{2}}\right) + \sum_{k=N}^{\infty} \left(\frac{1}{16k^{3}} + \frac{1}{4n^{4}k^{2}}\right) \\ = O\left(\frac{1}{n^{2}}\right) $$

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