3

$$ \lim_{n \rightarrow \infty} \sum_{k=1}^n \left( \sqrt{n^4+k}\ \sin\frac{2k\pi}{n} \right) = ? $$

I tried to transform it into a Riemann sum, to use Taylor-series of the sine function, to estimate, but nothing. Any help would be great.

Lacitek
  • 33
  • Not an answer, but a suggestion. Pair off values of $k$ and $k + n/2$ (this obviously works more smoothly when $n$ is even!). The sines are opposite, so the terms nearly cancel. Heuristically, this approach seems suggests a limit of $-1/4\pi$. – Dave Sep 06 '14 at 00:35
  • Numerical evaluations confirm @Dave's approach. – Lucian Sep 06 '14 at 00:47

2 Answers2

4

Hint: Factor $n^2$ outside the radical, and expand $\sqrt{1+\dfrac k{n^4}}$ into its binomial series $($the first two terms should suffice$)$, then use the fact that $\displaystyle\sum_1^n\sin\frac{2k\pi}n=0$, and $\displaystyle\int_0^1x\sin2\pi x~dx=-\dfrac1{2\pi}$.

Lucian
  • 48,334
  • 2
  • 83
  • 154
0

Certainly out of the scope of your problem $$\sum_{k=1}^n \sqrt{n^4+k}\ \sin\frac{2k\pi}{n}$$ is the imaginary part of $$e^{\frac{2 i \pi }{n}} \left(\Phi \left(e^{\frac{2 i \pi }{n}},-\frac{1}{2},n^4+1\right)-\Phi \left(e^{\frac{2 i \pi }{n}},-\frac{1}{2},n^4+n+1\right)\right)$$ where appears Lerch $\Phi$ function.

For large values of $n$, Taylor expansion leads to $$\sum_{k=1}^n \sqrt{n^4+k}\ \sin\frac{2k\pi}{n}=-\frac{1}{4 \pi }+\frac{\pi }{12 n^2}+\frac{1}{16 \pi n^3}+O\left(\left(\frac{1}{n}\right)^4\right)$$

Claude Leibovici
  • 260,315