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

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.

• $\cdots \leqslant \dfrac 1{k^2 \sqrt n}$ holds for $k \geqslant 2$, so it should be $$\frac 1n + \frac 1{\sqrt n}\sum_2^\infty \frac 1{k^2} \xrightarrow{n \to +\infty} 0.$$ – xbh Dec 3 '18 at 9:54

In your manipulations there is a mistake: note that for $$k\ge 2$$
$$\sqrt{n}\ge\sqrt[k]{n}\implies\frac1{k^2\sqrt[k]{n}}\ge\frac1{k^2\sqrt n}\tag1$$
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.