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.

  • $\begingroup$ $\cdots \leqslant \dfrac 1{k^2 \sqrt n}$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.