# Evaluating $\lim\limits_{n\to\infty} \dfrac{1}{n^3}\sum\limits_{\ell=1}^{n-1}\sqrt{(n^2-\ell^2)(n^2-(\ell-1)^2)}$

Along the way to proving a solution for this stubborn question of mine, I've come upon this expression which I would like to evaluate: $$\lim_{n\to\infty} \frac{1}{n^3}\sum_{\ell=1}^{n-1}\sqrt{(n^2-\ell^2)(n^2-(\ell-1)^2)}$$ Assuming consistency+correctness of the rest of my work, I would love for it to turn out that the limit is $\frac{2}{3}$, but to be honest I'm not certain of how to continue. I see no reason for there to be a nice closed form for the sum (and W|A appear to agree).

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## 1 Answer

For every $1\leqslant\ell\leqslant n-1$, $$n^2-\ell^2\leqslant\sqrt{(n^2-\ell^2)(n^2-(\ell-1)^2)}\leqslant n^2-(\ell-1)^2,$$ hence the sums $S_n$ you are interested in are such that $R_n\leqslant S_n\leqslant T_n$ for every $n\geqslant1$, with $$R_n=\frac1{n^3}\sum_{\ell=1}^{n-1}(n^2-\ell^2),\qquad T_n=\frac1{n^3}\sum_{\ell=0}^{n-2}(n^2-\ell^2).$$ The rest should be easy (and the limit is indeed $\frac23$).

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Excellent! And, thrilled to see that the solution is correct. I'm going to write up an answer to the old question now. – Eugene Shvarts Aug 13 '12 at 8:59
I wonder the question @did can not solve... – Seyhmus Güngören Aug 13 '12 at 9:19
Here is what I ended up doing with this. Thanks again! – Eugene Shvarts Aug 13 '12 at 9:41