I can't find the limit of this sequence series

Question

I can't solve this limit: $\lim_{n\to\infty}a_n=?$ when:

$a_n=\frac{1}{n}\sum_{k=1}^{n}(2k+1)*sin(\frac{k^2\pi}{n^2})$

Answer 1

$$a_n=\frac{1}{n}\sum_{k=1}^{n}(2k+1)*\sin(\frac{k^2\pi}{n^2})$$ $$=\sum_{k=1}^{n}(2\frac{k}{n}+\frac{1}{n})*\sin(\frac{k^2\pi}{n^2})$$ $$=n\sum_{k=1}^{n}\frac{1}{n}(2\frac{k}{n}\sin(\frac{k^2\pi}{n^2})) + \sum_{k=1}^{n}\frac{1}{n}\sin(\frac{k^2\pi}{n^2})$$

Now, as $n\to\infty$, we can convert the sums to an integral by taking $\frac{1}{n}=dx$ and $\frac{k}{n}=x$. We have:

$$a_n = n\int_{0}^{1}2x\sin(x^2\pi)dx + \int_o^1\sin(x^2\pi)dx$$ $$\approx \frac{2n}{\pi}+0.5 \approx \frac{2n}{\pi}$$ Thus, $\lim_{n\to\infty} a_n = \lim_{n\to\infty}\frac{2n}{\pi}=\infty$