$\lim\limits_{n\to\infty}n\big(\sum_{k=1}^n\frac{k^2}{n^3+kn}-\frac{1}{3}\big)$? 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
 A: There is another using generalized haromonic numbers since
$$\sum\limits_{k=1}^n\dfrac{k^2}{n^3+kn}=n^3 \left(H_{n^2+n}-H_{n^2}\right)+\frac{1}{2} \left(-2 n^2+n+1\right)$$
$$S_n=-\dfrac{1}{3}+\sum\limits_{k=1}^n\dfrac{k^2}{n^3+kn}=\frac{1}{6} \left(-6 n^2+3n+1\right)+n^3 \left(H_{n^2+n}-H_{n^2}\right)$$ Using asymptotics
$$S_n=\frac{1}{4 n}-\frac{2}{15 n^2}+\frac{1}{12 n^3}+O\left(\frac{1}{n^4}\right)$$
$$nS_n=\frac{1}{4 }-\frac{2}{15 n}+\frac{1}{12 n^2}+O\left(\frac{1}{n^3}\right)$$ and we do not agree !
Try with $n=10$; the exact value is
$$10 S_{10}=10 \times \frac{140325051799081}{5909102214621606}\sim 0.237473$$ while the approximation gives $\frac{19}{80} \sim 0.237500$.
A: Although $\sum\frac{k^2}{n^3+kn}$ and $\sum\frac{k^2}{n^3}$ have the same limit, they differ to $O(1/n)$. So when they are multiplied by $n$, they give different limits
A: 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,
$$ \frac{1}{2}-\frac{1}{4}=\frac{1}{4}$$
A: Your approach can be fixed easily. You have to evaluate the limit of $n(S_n-(1/3))$ and you have also shown that $$\frac{n} {n+1}R_n<S_n<R_n$$ where $R_n=n^{-1}\sum_{k=1}^{n}f(k/n)$ is the Riemann sum for $f(x) =x^2$ over $[0,1]$.
It is well known that $$n\left(R_n-\int_{0}^{1}f(x)\,dx\right)\to\frac{f(1)-f(0)}{2}=\frac{1}{2}\tag{1}$$ Our job is complete if we can evaluate the limit $L$ of $n(S_n-R_n)$ and our desired limit will be $L+1/2$.
But $$n(S_n-R_n) =-\sum_{k=1}^{n}\frac{k/n^2}{1+(k/n^2)}\cdot\frac{k^2}{n^2}$$ The sum on right can again be squeezed (as in question) to get the limit $L=-1/4$. The desired limit is thus $1/4$.
Moral of the Story: Do not deviate even an iota from the hypotheses and conclusion of a theorem. If you do that try to do further analysis and prove the desired changes. The result (equation $(1)$ above) you use deals with a specific Riemann sum and that can't be replaced by any similar looking sum (even having same limit).
