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).


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 $\frac23$).

| cite | improve this answer | |
  • 1
    $\begingroup$ Excellent! And, thrilled to see that the solution is correct. I'm going to write up an answer to the old question now. $\endgroup$ – Eugene Shvarts Aug 13 '12 at 8:59
  • 2
    $\begingroup$ I wonder the question @did can not solve... $\endgroup$ – Seyhmus Güngören Aug 13 '12 at 9:19
  • $\begingroup$ Here is what I ended up doing with this. Thanks again! $\endgroup$ – Eugene Shvarts Aug 13 '12 at 9:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.