How to find limit of $\sum\limits_{k = 1}^{n - 1}(1 + \frac{k}{n})sin\frac{k\pi}{n^2}$

Question

What i should to do then see sums like this?

@Dr.SonnhardGraubner Does WA provide guidance on methodologies to attack similar problems? The OP is not simply seeking a value to the limit. The OP is requesting possible ways forward. Would you please refrain from reliance on computer algorithms to replace actual thought. ;-) — Mark Viola, May 15 '17 at 19:34

score 3 · Accepted Answer · answered May 15 '17 at 19:01

3

Using Taylor Lagrange inequality we have

$$\left\vert\sin\left(\frac{k\pi}{n^2}\right)-\frac{k\pi}{n^2}\right\vert\le \frac{M}{n^3}$$ so with this inequality we prove that the desired limit is equal to

$$\lim_{n\to\infty}\frac1n\sum_{k=1}^{n-1}\left(1+\frac kn\right)\frac{k\pi}n=\int_0^1(1+x)\pi xdx$$ using the Riemann sum.

answered May 15 '17 at 19:01

user296113

7,570

1

(+1) This is solid. HERE is a related problem. – Mark Viola May 15 '17 at 19:35

How to find limit of $\sum\limits_{k = 1}^{n - 1}(1 + \frac{k}{n})sin\frac{k\pi}{n^2}$

1 Answers1