Computing the limit $\lim_{n \to \infty} \sum_{k=1}^{n} \sqrt{n^4+k}\sin \frac{2k \pi}{n}$

Asked Apr 01 '15 at 12:51

Active Apr 01 '15 at 13:47

Viewed 94 times

I'm pretty sure this is a Riemann sum, but I can't solve it. I tried to write it as the integral of a function which equals the product of $\sin(2 \pi x)$ and something, but so far it didn't work.

$$\lim_{n \to \infty} \sum_{k=1}^{n} \sqrt{n^4+k}\sin \frac{2k \pi}{n}$$

edited Apr 01 '15 at 13:47

Pedro

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asked Apr 01 '15 at 12:51

Alexander Smith

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$L~=~-\dfrac1{4\pi}$ – Lucian Apr 01 '15 at 13:42
@Lucian: I just determined that using Richardson extrapolation and the ISC, but I was trying to figure out a way to give a hint without simply stating the value. :) I suspect that Alexander would like a proof, not just the value of the limit, but I guess knowing the answer can be helpful in figuring out if your proof is valid. – PM 2Ring Apr 01 '15 at 13:47
@PM2Ring: Oddly enough, the series containing $n^2$ instead of $\sqrt{n^4+k}$ converges to $0$. – Lucian Apr 01 '15 at 13:52
Let us continue this discussion in chat. – PM 2Ring Apr 02 '15 at 12:39

Computing the limit $\lim_{n \to \infty} \sum_{k=1}^{n} \sqrt{n^4+k}\sin \frac{2k \pi}{n}$

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