vrotterigzl

Answered

2022-06-24

I would like to evaluate:
$\underset{n\to \mathrm{\infty }}{lim}\frac{1}{{n}^{3}}\sum _{\ell =1}^{n-1}\sqrt{\left({n}^{2}-{\ell }^{2}\right)\left({n}^{2}-\left(\ell -1{\right)}^{2}\right)}$

Answer & Explanation

Zayden Andrade

Expert

2022-06-25Added 22 answers

For every $1⩽\ell ⩽n-1$
${n}^{2}-{\ell }^{2}⩽\sqrt{\left({n}^{2}-{\ell }^{2}\right)\left({n}^{2}-\left(\ell -1{\right)}^{2}\right)}⩽{n}^{2}-\left(\ell -1{\right)}^{2},$
hence the sums ${S}_{n}$ you are interested in are such that ${R}_{n}⩽{S}_{n}⩽{T}_{n}$ for every $n⩾1$, with
${R}_{n}=\frac{1}{{n}^{3}}\sum _{\ell =1}^{n-1}\left({n}^{2}-{\ell }^{2}\right),\phantom{\rule{2em}{0ex}}{T}_{n}=\frac{1}{{n}^{3}}\sum _{\ell =0}^{n-2}\left({n}^{2}-{\ell }^{2}\right).$
The rest should be easy (and the limit is indeed $\frac{2}{3}$)

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