2
$\begingroup$

I have to calculate this limit using Riemann sums: $$\lim _{n\to \infty }\left[\left(1+\frac{1}{n}\right)\sin \left(\frac{\pi }{n^2}\right)+\left(1+\frac{2}{n}\right)\sin \:\left(\frac{2\pi \:}{n^2}\right)+...+\left(1+\frac{n-1}{n}\right)\sin \:\left(\frac{\left(n-1\right)\pi \:}{n^2}\right)\right]$$ and $$\lim _{n\to \infty }\left[\sum _{k=1}^{n-1}\left(1+\frac{k}{n}\right)\sin \left(\frac{k\pi }{n^2}\right)\right]$$

I've done quite a few of this type of limits, but I haven't yet figured out what function $f$ to use with the Riemann sum for this particular one. Could I have some hints on how to find that function? Thank you.

$\endgroup$
2
$\begingroup$

Hint. Note that, for $x\geq 0$ $$x-\frac{x^3}{6}\leq \sin(x)\leq x.$$ By using the upper bound, it follows that, as $n\to\infty$, $$\sum _{k=1}^{n-1}\left(1+\frac{k}{n}\right)\sin \left(\frac{k\pi }{n^2}\right)\leq \frac{\pi}{n}\sum _{k=1}^{n-1}\left(1+\frac{k}{n}\right)\left(\frac{k }{n}\right)\to\pi\int_0^1(1+x)xdx=\frac{5\pi}{6}.$$ Now for the other side, use the lower bound, and consider that $$0\leq \sum _{k=1}^{n-1}\left(1+\frac{k}{n}\right)\left(\frac{k\pi }{n^2}\right)^3=\frac{\pi^3}{n^2}\cdot\frac{1}{n}\sum _{k=1}^{n-1}\left(1+\frac{k}{n}\right)\left(\frac{k }{n}\right)^3.$$

$\endgroup$
  • $\begingroup$ Do we need to show LHR either ? or it is obvious but I couldn't see $\endgroup$ – haqnatural Nov 27 '17 at 19:43
  • $\begingroup$ Yes, you need to show it.See the final part of my answer. $\endgroup$ – Robert Z Nov 27 '17 at 19:52

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.