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Question Number 205307 by universe last updated on 15/Mar/24

lim_(n→∞) ((⌊a⌋+⌊2a⌋+...+⌊na⌋)/n^2 ) where a∈R and ⌊x⌋ is the floor of x ∈ R

$$\:\:\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\lfloor{a}\rfloor+\lfloor\mathrm{2}{a}\rfloor+...+\lfloor{na}\rfloor}{{n}^{\mathrm{2}} }\:\mathrm{where}\:{a}\in\mathbb{R} \\ $$$$\:\:\:\mathrm{and}\:\lfloor{x}\rfloor\:\mathrm{is}\:\mathrm{the}\:\mathrm{floor}\:\mathrm{of}\:\mathrm{x}\:\in\:\mathbb{R} \\ $$

Commented by Frix last updated on 15/Mar/24

Just guessing: −∞ for a<0 (a/2) for a≥0

$$\mathrm{Just}\:\mathrm{guessing}: \\ $$$$−\infty\:\mathrm{for}\:{a}<\mathrm{0} \\ $$$$\frac{{a}}{\mathrm{2}}\:\mathrm{for}\:{a}\geqslant\mathrm{0} \\ $$

Answered by Mathspace last updated on 15/Mar/24

[a]≤a<[a]+1 ⇒[a]≤a eta−1<[a] ⇒ a−1<[a]≤a ka−1<[ka]≤ka ⇒ Σ_(k=1) ^n (ka−1)<Σ_(k=1) ^n [ka]≤Σ_(k=1) ^n ka ⇒a.((n(n+1))/2)−n<Σ_(k=1) ^n [ka]≤a((n(n+1))/2) ⇒((n(n+1))/n^2 )a−(1/n)<((Σ_(k=1) ^n [ka])/n^2 )≤((n(n+1))/(2n^2 ))a ona lim_(n→+∞) ((n(n+1))/(2n^2 ))a−(1/n)=(a/2) lim_(n→+∞) ((n(n+1)a)/(2n^2 ))=(a/2) ⇒ lim_(n→+∞) ((Σ_(k=1) ^n [ka])/n^2 )=(a/2)

$$\left[{a}\right]\leqslant{a}<\left[{a}\right]+\mathrm{1}\:\Rightarrow\left[{a}\right]\leqslant{a}\:\:{eta}−\mathrm{1}<\left[{a}\right]\:\Rightarrow \\ $$$${a}−\mathrm{1}<\left[{a}\right]\leqslant{a} \\ $$$${ka}−\mathrm{1}<\left[{ka}\right]\leqslant{ka}\:\Rightarrow \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \left({ka}−\mathrm{1}\right)<\sum_{{k}=\mathrm{1}} ^{{n}} \left[{ka}\right]\leqslant\sum_{{k}=\mathrm{1}} ^{{n}} {ka} \\ $$$$\Rightarrow{a}.\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}−{n}<\sum_{{k}=\mathrm{1}} ^{{n}} \left[{ka}\right]\leqslant{a}\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}} \\ $$$$\Rightarrow\frac{{n}\left({n}+\mathrm{1}\right)}{{n}^{\mathrm{2}} }{a}−\frac{\mathrm{1}}{{n}}<\frac{\sum_{{k}=\mathrm{1}} ^{{n}} \left[{ka}\right]}{{n}^{\mathrm{2}} }\leqslant\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}{n}^{\mathrm{2}} }{a} \\ $$$${ona}\:{lim}_{{n}\rightarrow+\infty} \frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}{n}^{\mathrm{2}} }{a}−\frac{\mathrm{1}}{{n}}=\frac{{a}}{\mathrm{2}} \\ $$$${lim}_{{n}\rightarrow+\infty} \frac{{n}\left({n}+\mathrm{1}\right){a}}{\mathrm{2}{n}^{\mathrm{2}} }=\frac{{a}}{\mathrm{2}}\:\Rightarrow \\ $$$${lim}_{{n}\rightarrow+\infty} \frac{\sum_{{k}=\mathrm{1}} ^{{n}} \left[{ka}\right]}{{n}^{\mathrm{2}} }=\frac{{a}}{\mathrm{2}} \\ $$

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