The problem:
$$
\lim_{n \to \infty} \sum_{i=0}^n \frac{i\pi}{n^2}\sin{(\frac{i^2\pi^2}{4n^2})}
$$
My attempt:
Removed constants
$$
\pi\lim_{n \to \infty} \frac{1}{n^2}\sum_{i=0}^n i\sin{(\frac{i^2\pi^2}{4n^2})}
$$
Considered l'Hospital's Rule since $n^2 \to \infty$ and $\sum_{i=0}^n i\sin{(\frac{i^2\pi^2}{4n^2})}$ does not converge by the limit convergence test.
L'Hospital's Rule gives
$$
\pi\lim_{n \to \infty} \frac{1}{2n}\sum_{i=0}^n i\sin{(\frac{i^2\pi^2}{4n^2})}(-2)\frac{i^2\pi^2}{4n^3}
$$
$$
=\frac{-\pi^3}{4}\lim_{n \to \infty} \frac{1}{n^4} \sum_{i=0}^n i^3\cos{\frac{i^2\pi^2}{4n^2}}
$$
Which once again has an indeterminant form. I don't see myself getting anywhere with l'Hospital's Rule again, and I'm not really sure what I can do. Any tips would help tons! Thanks.
Best Answer
First, we may notice that we can make the sum start at $i = 1$ since the first term is $0$.
Let $f : x \in [0, \pi] \mapsto x \sin\left(\frac{x^2}{4}\right)$. This function is continuous, hence: $$\frac{\pi}{n} \sum_{i = 1}^n f\left(\frac{i\pi}{n}\right) \xrightarrow[n \to \infty]{} \int_0^\pi f(t) \mathrm{d}t$$ Yet: $\frac{\pi}{n} \sum_{i = 1}^n f\left(\frac{i\pi}{n}\right) = \pi\sum_{i=1}^n \frac{i\pi}{n^2}\sin\left(\frac{i^2\pi^2}{4n^2}\right)$, thus the limit you need will be: $$\frac1{\pi}\int_0^\pi f(t)\mathrm{d}t = \frac1{\pi}\int_0^\pi t\sin\left(\frac{t^2}{4}\right)\mathrm{d}t$$ Fortunaetly this integral can be simplified manually: $$\begin{split} \int_0^\pi t\sin\left(\frac{t^2}{4}\right)\mathrm{d}t &= 2 \int_0^\pi \frac{2t}{4}\sin\left(\frac{t^2}{4}\right)\mathrm{d}t \\ &= 2 \left[-\cos\left(\frac{t^2}{4}\right)\right]_{t = 0}^{t = \pi} \\ &= 2 \left(1 - \cos\left(\frac{\pi^2}{4}\right)\right) \\ &= 4\sin^2\left(\frac{\pi^2}{8}\right)\end{split}$$ where the last line uses the identity $\sin^2(\theta) = \frac{1 - \cos(2\theta)}{2}$.
Therefore, the result is: $$\lim_{n \to \infty} \sum_{i=1}^n \frac{i\pi}{n^2}\sin\left(\frac{i^2\pi^2}{4n^2}\right) \xrightarrow[n \to \infty]{} \frac{4}{\pi}\sin^2\left(\frac{\pi^2}{8}\right)$$