Let x be any positive real number, and define a sequence [an] by an=[x]+[2x]+[3x]+…+[nx]n2 where [x] is the largest integer less than or equal to x. Prove that limn→∞an=x2

Since [x]=x+O(1) we see that [x]+[2x]+…+[nx]n2=x+2x+…+nxn2+O(1n) Summing, since x is a fixed constant, this becomes x2+O(1n) which converges to x2 as n→∞

So, you get x+2x+3x+…+nx−nn2≤an≤x+2x+3x+…+nxn2 for all n∈N. Its

By definition of the floor function, kx−1≤[kx]≤kx Then summing on k from 1 to n, n(n+1)2x−n≤∑k=1n[kx]≤(n+1)n2x Dividing by n2 and taking the limit, x2≤1n2∑k=1∞≤x2

Since [x]=x+O(1) we see that [x]+[2x]+…+[nx]n2=x+2x+…+nxn2+O(1n) Summing, since x is a fixed constant, this becomes x2+O(1n) which converges to x2 as n→∞

So, you get x+2x+3x+…+nx−nn2≤an≤x+2x+3x+…+nxn2 for all n∈N. Its

By definition of the floor function, kx−1≤[kx]≤kx Then summing on k from 1 to n, n(n+1)2x−n≤∑k=1n[kx]≤(n+1)n2x Dividing by n2 and taking the limit, x2≤1n2∑k=1∞≤x2

Explained: "Let x be any positive real number, and..."

maliaseth0

Answered question

2022-01-22

Let x be any positive real number, and define a sequence

[a_{n}]

a_{n} = \frac{[x] + [2 x] + [3 x] + \dots + [n x]}{n^{2}}

where

[x]

is the largest integer less than or equal to x. Prove that

lim_{n \to \infty} a_{n} = \frac{x}{2}

Answer & Explanation

euromillionsna

Beginner2022-01-23Added 16 answers

Since

[x] = x + O (1)

we see that

\frac{[x] + [2 x] + \dots + [n x]}{n^{2}} = \frac{x + 2 x + \dots + n x}{n^{2}} + O (\frac{1}{n})

Summing, since x is a fixed constant, this becomes

\frac{x}{2} + O (\frac{1}{n})

which converges to

\frac{x}{2}

n \to \infty

kumewekwah0

Beginner2022-01-24Added 14 answers

So, you get

\frac{x + 2 x + 3 x + \dots + n x - n}{n^{2}} \leq a_{n} \leq \frac{x + 2 x + 3 x + \dots + n x}{n^{2}}

for all

n \in N

. Its

RizerMix

Expert2022-01-27Added 656 answers

By definition of the floor function,

k x - 1 \leq [k x] \leq k x

Then summing on k from 1 to n,

\frac{n (n + 1)}{2} x - n \leq \sum_{k = 1}^{n} [k x] \leq \frac{(n + 1) n}{2} x

Dividing by

n^{2}

and taking the limit,

\frac{x}{2} \leq \frac{1}{n^{2}} \sum_{k = 1}^{\infty} \leq \frac{x}{2}

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