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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

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  • $\begingroup$ Good question. And nice attempt (even though final answer is not correct). +1 $\endgroup$ – Paramanand Singh Jan 23 at 13:24
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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).

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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 $\frac{19}{80} \sim 0.237500$.

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  • $\begingroup$ thank you,so is my answer wrong? $\endgroup$ – Hilbert1994 Jan 23 at 9:56
  • $\begingroup$ @夜愿1998. It seems that your answer is not correct. $\endgroup$ – Claude Leibovici Jan 23 at 10:25
  • $\begingroup$ thank you,I want to know which step is wrong $\endgroup$ – Hilbert1994 Jan 23 at 10:28
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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

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  • $\begingroup$ Thank you. How can I improve my calculation $\endgroup$ – Hilbert1994 Jan 23 at 10:56
  • $\begingroup$ Perhaps multiply top and bottom by $n^2-k$ because the new denominator is between $n^5$ and $n^5-n^3$ $\endgroup$ – Empy2 Jan 23 at 11:23
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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, $$ \frac{1}{2}-\frac{1}{4}=\frac{1}{4}$$

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  • $\begingroup$ I have a question. Is my asymptotic formula not precise enough? $\endgroup$ – Hilbert1994 Jan 23 at 11:36
  • $\begingroup$ It is precise (if you put the plus instead of minus in the second term), but you used only two first terms of it in your calculations (which are enough to solve this problem). In fact there are other terms (which decline as powers of $\frac{1}{n}$);$B_k$ means Bernoulli coefficients. $\endgroup$ – Svyatoslav Jan 23 at 11:43
  • $\begingroup$ Sorry, I don't quite understand you. Which equation do you mean? $\endgroup$ – Hilbert1994 Jan 23 at 11:49
  • $\begingroup$ and $\lim\limits_{n\to\infty}\dfrac{1}{n}\sum\limits_{k=1}^n\dfrac{(\frac{k}{n})^2}{1+\frac{k}{n^2}}$ can be $\int_0^1\frac{t^2}{1+\frac{1}{n}t}dt$? $\endgroup$ – Hilbert1994 Jan 23 at 11:52
  • $\begingroup$ I mean the approximation of the sum which you use: $\lim\limits_{n\to\infty}n\left(\int_0^1f(x)\mathrm{d}x-\frac{1}{n}\sum\limits_{k=1}^{n}f\left(\frac{k}{n}\right)\right)=\frac{f(0)-f(1)}{2}$ $\endgroup$ – Svyatoslav Jan 23 at 11:52

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