$\lim\limits_{n\to\infty}n\big(\sum_{k=1}^n\frac{k^2}{n^3+kn}-\frac{1}{3}\big)$? calculate $$\lim\limits_{n\to\infty}n\left(\sum\limits_{k=1}^n\dfrac{k^2}{n^3+kn}-\dfrac{1}{3}\right).$$
I got it $$\lim\limits_{n\to\infty}\sum\limits_{k=1}^n\dfrac{k^2}{n^3+kn}=\lim\limits_{n\to\infty}\dfrac{1}{n}\sum\limits_{k=1}^n\dfrac{(\frac{k}{n})^2}{1+\frac{k}{n^2}}.$$
Use Squeeze theorem we have
$$\frac{1}{n+1}\sum\limits_{k=1}^n(\frac{k}{n})^2<\dfrac{1}{n}\sum\limits_{k=1}^n\dfrac{(\frac{k}{n})^2}{1+\frac{k}{n^2}}<\dfrac{1}{n}\sum\limits_{k=1}^n(\frac{k}{n})^2$$
So $$\lim\limits_{n\to\infty}\sum\limits_{k=1}^n\dfrac{k^2}{n^3+kn}=\int_0^1x^2\mathrm{d}x=\frac{1}{3}.$$
Use $$\lim\limits_{n\to\infty}n\left(\int_0^1f(x)\mathrm{d}x-\frac{1}{n}\sum\limits_{k=1}^{n}f\left(\frac{k}{n}\right)\right)=\frac{f(0)-f(1)}{2}.$$
Hence $$\lim\limits_{n\to\infty}n\left(\sum\limits_{k=1}^n\dfrac{k^2}{n^3+kn}-\dfrac{1}{3}\right)=\frac{1}{2}.$$
If our method is correct, is there any other way to solve this problem? Thank you

In fact you are very close to the right answer. I would suggest you check Euler - Maclaurin formula first - plus, not minus in the second term, though it does not change the result that you got $(+\frac{1}{2})$: $$\sum\limits_{k=1}^nf(\frac{k}{n})=n\int_1^nf(t)dt+\frac{1}{2}\left(f(\frac{1}{n})+f(1)\right)+\sum\limits_{k=2}^{\infty}(\frac{1}{n})^{k-1}\frac{B_k}{k!}\left(f^{(k-1)}(1)-f^{(k-1)}(\frac{1}{n})\right)$$
Next, the integral in fact is $$n\int_0^1\frac{t^2}{1+\frac{1}{n}t}dt=n\int_0^1t^2dt-\int_0^1t^3dt+O(\frac{1}{n})$$ the second term here gives you additionally $-\frac{1}{4}$.
All together,
$$ \frac{1}{2}-\frac{1}{4}=\frac{1}{4}$$
