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lim-n-a-2a-na-n-2-where-a-R-and-x-is-the-floor-of-x-R-




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
limna+2a++nan2whereaRandxisthefloorofxR
Commented by Frix last updated on 15/Mar/24
Just guessing:  −∞ for a<0  (a/2) for a≥0
Justguessing:fora<0a2fora0
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)
[a]a<[a]+1[a]aeta1<[a]a1<[a]aka1<[ka]kak=1n(ka1)<k=1n[ka]k=1nkaa.n(n+1)2n<k=1n[ka]an(n+1)2n(n+1)n2a1n<k=1n[ka]n2n(n+1)2n2aonalimn+n(n+1)2n2a1n=a2limn+n(n+1)a2n2=a2limn+k=1n[ka]n2=a2

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