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Find $$\lim\limits_{n \to \infty} \sum\limits_{k=1}^{\infty}\frac{1}{k^{2}\sqrt[k]{n}}\sin^{2}\left(\frac{n \pi}{k}\right)$$

This is the first time that I am operating with $\lim_{n\to \infty}\lim_{k \to \infty}$ so I am unsure. My first idea would be to look at:

$\frac{1}{k^{2}\sqrt[k]{n}}\sin^{2}(\frac{n \pi}{k})$ where $n \in \mathbb N$ is constant.

$\frac{1}{k^{2}\sqrt[k]{n}}\sin^{2}(\frac{n \pi}{k})\leq \frac{1}{k^{2}\sqrt[k]{n}}\leq\frac{1}{k^{2}\sqrt{n}}$

and $\sum_{k=1}^{\infty}\frac{1}{k^{2}\sqrt{n}}=\frac{1}{\sqrt{n}}\sum_{k=1}^{\infty}\frac{1}{k^{2}}$

and we know $\sum_{k=1}^{\infty}\frac{1}{k^{2}} < \infty$ and taking $n \to \infty$ we get

$\lim_{n\to \infty}\frac{1}{\sqrt{n}}\sum_{k=1}^{\infty}\frac{1}{k^{2}}=0=\lim_{n \to \infty} \sum_{k=1}^{\infty}\frac{1}{k^{2}\sqrt[k]{n}}\sin^{2}(\frac{n \pi}{k})$

I assume this is incorrect. Help/Corrections would be greatly appreciated.

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  • $\begingroup$ $\cdots \leqslant \dfrac 1{k^2 \sqrt n}$ holds for $k \geqslant 2$, so it should be $$ \frac 1n + \frac 1{\sqrt n}\sum_2^\infty \frac 1{k^2} \xrightarrow{n \to +\infty} 0.$$ $\endgroup$ – xbh Dec 3 '18 at 9:54
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In your manipulations there is a mistake: note that for $k\ge 2$

$$\sqrt{n}\ge\sqrt[k]{n}\implies\frac1{k^2\sqrt[k]{n}}\ge\frac1{k^2\sqrt n}\tag1$$

A way to solve this limit is using the Weierstrass M-test and the properties of uniform convergence of series.

Note that for all $k\in\Bbb N_{\ge 1}$ and $x\ge 1$ it holds that $\sqrt[k]{x}\ge 1$, consequently

$$\frac1{k^2}\ge\frac1{k^2\sqrt[k]{n}}\ge\frac1{k^2\sqrt[k]{n}}\sin^2(n \pi/k)\tag2$$

Hence by the M-test the series $\sum_{k=1}^\infty f_k(x)$, for $f_k(x):=\frac1{k^2\sqrt[k]{x}}\sin^2(x \pi/k)$, converges absolutely and uniformly for $x\ge 1$, so we can exchange limit and summation sign to find that the limit that we want to evaluate is indeed zero.

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