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$limlimits_{ntoinfty}{[x]+[2^2x]+dots +[n^2x]over n^3}$

3

1

$begingroup$

Find the limit, $$lim_{ntoinfty}{[x]+[2^2x]+dots +[n^2x]over n^3}$$

Where $[.]$ denotes the integral part of $.$?

Efforts: If $x$ is integer,
$$lim_{ntoinfty}{[x]+[2^2x]+dots +[n^2x]over n^3}$$ $$lim_{ntoinfty}{x(1+2^2+dots +n^2)over n^3}$$

We know $sumlimits_{i=1}^ni^2=(n+1)(2n+1)n/6$

Therefore we get that limit is equal to $x/3$.

How to solve it for non integral values.
Thanks in advance.

real-analysis limits

share|cite|improve this question

edited Mar 24 at 10:59

rtybase

11.6k31534

asked Mar 24 at 10:22

StammeringMathematicianStammeringMathematician

2,8171324

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  You can try to use $x-1 leq [x] leq x$ and squeeze out the limit.
  $endgroup$
  – Winther
  Mar 24 at 10:31
  
  
  

  
  
  
  

add a comment | 

3

1

$begingroup$

Find the limit, $$lim_{ntoinfty}{[x]+[2^2x]+dots +[n^2x]over n^3}$$

Where $[.]$ denotes the integral part of $.$?

Efforts: If $x$ is integer,
$$lim_{ntoinfty}{[x]+[2^2x]+dots +[n^2x]over n^3}$$ $$lim_{ntoinfty}{x(1+2^2+dots +n^2)over n^3}$$

We know $sumlimits_{i=1}^ni^2=(n+1)(2n+1)n/6$

Therefore we get that limit is equal to $x/3$.

How to solve it for non integral values.
Thanks in advance.

real-analysis limits

share|cite|improve this question

edited Mar 24 at 10:59

rtybase

11.6k31534

asked Mar 24 at 10:22

StammeringMathematicianStammeringMathematician

2,8171324

$endgroup$

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  You can try to use $x-1 leq [x] leq x$ and squeeze out the limit.
  $endgroup$
  – Winther
  Mar 24 at 10:31
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Find the limit, $$lim_{ntoinfty}{[x]+[2^2x]+dots +[n^2x]over n^3}$$

Where $[.]$ denotes the integral part of $.$?

Efforts: If $x$ is integer,
$$lim_{ntoinfty}{[x]+[2^2x]+dots +[n^2x]over n^3}$$ $$lim_{ntoinfty}{x(1+2^2+dots +n^2)over n^3}$$

We know $sumlimits_{i=1}^ni^2=(n+1)(2n+1)n/6$

Therefore we get that limit is equal to $x/3$.

How to solve it for non integral values.
Thanks in advance.

real-analysis limits

share|cite|improve this question

edited Mar 24 at 10:59

rtybase

11.6k31534

asked Mar 24 at 10:22

StammeringMathematicianStammeringMathematician

2,8171324

$endgroup$

share|cite|improve this question

edited Mar 24 at 10:59

rtybase

11.6k31534

asked Mar 24 at 10:22

StammeringMathematicianStammeringMathematician

2,8171324

share|cite|improve this question

edited Mar 24 at 10:59

rtybase

11.6k31534

asked Mar 24 at 10:22

StammeringMathematicianStammeringMathematician

2,8171324

edited Mar 24 at 10:59

rtybase

11.6k31534

edited Mar 24 at 10:59

rtybase

11.6k31534

asked Mar 24 at 10:22

StammeringMathematicianStammeringMathematician

2,8171324

asked Mar 24 at 10:22

StammeringMathematicianStammeringMathematician

2,8171324

2 Answers
2

active

oldest

votes

2

$begingroup$

Applying Stolz–Cesàro theorem with
$a_n=sumlimits_{k=1}^n [xcdot k^2]$ and
$b_n=n^3$ (which is strictly monotone and divergent)
$$frac{a_{n+1}-a_n}{b_{n+1}-b_n}=
frac{[x(n+1)^2]}{(n+1)^3-n^3}=
frac{[x(n+1)^2]}{(n+1)^2+(n+1)n+n^2}$$
and we have (using inequality @Winther suggested in the comments $x-1leq[x]<x$ and squeeze theorem)
$$frac{x(n+1)^2-1}{3(n+1)^2}leq
frac{[x(n+1)^2]}{3(n+1)^2}<
frac{a_{n+1}-a_n}{b_{n+1}-b_n}<
frac{[x(n+1)^2]}{3n^2}<
frac{x(n+1)^2}{3n^2}$$
as a result
$$limlimits_{nrightarrowinfty}frac{a_{n+1}-a_n}{b_{n+1}-b_n}=frac{x}{3}$$
and finally
$$limlimits_{nrightarrowinfty}frac{a_n}{b_n}=
limlimits_{nrightarrowinfty}frac{sumlimits_{k=1}^n [xcdot k^2]}{n^3}=
frac{x}{3}$$

share|cite|improve this answer

edited Mar 24 at 11:00

answered Mar 24 at 10:52

rtybasertybase

11.6k31534

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  simply beautiful :)
  $endgroup$
  – StammeringMathematician
  Mar 24 at 10:59
  

  
  
  
  

add a comment | 

2

$begingroup$

See the difference between this one and
$frac{x+2^2x+3^2x+...+n^2 x}{n^3}$ is $frac{{x}+{2^2x}+{3^2x}+...+{n^2 x}}{n^3}$, which is smaller than $frac{n^2}{n^3}=frac 1n to 0$; here ${y}=y-[y]$. So your expression has the same limit $frac 13$ as $frac{x+2^2x+3^2x+...+n^2 x}{n^3}$.

share|cite|improve this answer

answered Mar 24 at 10:41

Yu DingYu Ding

7287

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Probably a typo: makes more sense to say it's less than $frac{n}{n^3}$ instead of $frac{n^2}{n^3}$ as there are $n$ terms in the numerator each of which is less than $1$.
  $endgroup$
  – Winther
  Mar 24 at 15:04
  

  
  
  
  

add a comment | 

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2

$begingroup$

Applying Stolz–Cesàro theorem with
$a_n=sumlimits_{k=1}^n [xcdot k^2]$ and
$b_n=n^3$ (which is strictly monotone and divergent)
$$frac{a_{n+1}-a_n}{b_{n+1}-b_n}=
frac{[x(n+1)^2]}{(n+1)^3-n^3}=
frac{[x(n+1)^2]}{(n+1)^2+(n+1)n+n^2}$$
and we have (using inequality @Winther suggested in the comments $x-1leq[x]<x$ and squeeze theorem)
$$frac{x(n+1)^2-1}{3(n+1)^2}leq
frac{[x(n+1)^2]}{3(n+1)^2}<
frac{a_{n+1}-a_n}{b_{n+1}-b_n}<
frac{[x(n+1)^2]}{3n^2}<
frac{x(n+1)^2}{3n^2}$$
as a result
$$limlimits_{nrightarrowinfty}frac{a_{n+1}-a_n}{b_{n+1}-b_n}=frac{x}{3}$$
and finally
$$limlimits_{nrightarrowinfty}frac{a_n}{b_n}=
limlimits_{nrightarrowinfty}frac{sumlimits_{k=1}^n [xcdot k^2]}{n^3}=
frac{x}{3}$$

share|cite|improve this answer

edited Mar 24 at 11:00

answered Mar 24 at 10:52

rtybasertybase

11.6k31534

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  simply beautiful :)
  $endgroup$
  – StammeringMathematician
  Mar 24 at 10:59
  

  
  
  
  

add a comment | 

2

$begingroup$

Applying Stolz–Cesàro theorem with
$a_n=sumlimits_{k=1}^n [xcdot k^2]$ and
$b_n=n^3$ (which is strictly monotone and divergent)
$$frac{a_{n+1}-a_n}{b_{n+1}-b_n}=
frac{[x(n+1)^2]}{(n+1)^3-n^3}=
frac{[x(n+1)^2]}{(n+1)^2+(n+1)n+n^2}$$
and we have (using inequality @Winther suggested in the comments $x-1leq[x]<x$ and squeeze theorem)
$$frac{x(n+1)^2-1}{3(n+1)^2}leq
frac{[x(n+1)^2]}{3(n+1)^2}<
frac{a_{n+1}-a_n}{b_{n+1}-b_n}<
frac{[x(n+1)^2]}{3n^2}<
frac{x(n+1)^2}{3n^2}$$
as a result
$$limlimits_{nrightarrowinfty}frac{a_{n+1}-a_n}{b_{n+1}-b_n}=frac{x}{3}$$
and finally
$$limlimits_{nrightarrowinfty}frac{a_n}{b_n}=
limlimits_{nrightarrowinfty}frac{sumlimits_{k=1}^n [xcdot k^2]}{n^3}=
frac{x}{3}$$

share|cite|improve this answer

edited Mar 24 at 11:00

answered Mar 24 at 10:52

rtybasertybase

11.6k31534

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  simply beautiful :)
  $endgroup$
  – StammeringMathematician
  Mar 24 at 10:59
  

  
  
  
  

add a comment | 

$begingroup$

Applying Stolz–Cesàro theorem with
$a_n=sumlimits_{k=1}^n [xcdot k^2]$ and
$b_n=n^3$ (which is strictly monotone and divergent)
$$frac{a_{n+1}-a_n}{b_{n+1}-b_n}=
frac{[x(n+1)^2]}{(n+1)^3-n^3}=
frac{[x(n+1)^2]}{(n+1)^2+(n+1)n+n^2}$$
and we have (using inequality @Winther suggested in the comments $x-1leq[x]<x$ and squeeze theorem)
$$frac{x(n+1)^2-1}{3(n+1)^2}leq
frac{[x(n+1)^2]}{3(n+1)^2}<
frac{a_{n+1}-a_n}{b_{n+1}-b_n}<
frac{[x(n+1)^2]}{3n^2}<
frac{x(n+1)^2}{3n^2}$$
as a result
$$limlimits_{nrightarrowinfty}frac{a_{n+1}-a_n}{b_{n+1}-b_n}=frac{x}{3}$$
and finally
$$limlimits_{nrightarrowinfty}frac{a_n}{b_n}=
limlimits_{nrightarrowinfty}frac{sumlimits_{k=1}^n [xcdot k^2]}{n^3}=
frac{x}{3}$$

share|cite|improve this answer

edited Mar 24 at 11:00

answered Mar 24 at 10:52

rtybasertybase

11.6k31534

$endgroup$

share|cite|improve this answer

edited Mar 24 at 11:00

answered Mar 24 at 10:52

rtybasertybase

11.6k31534

answered Mar 24 at 10:52

rtybasertybase

11.6k31534

answered Mar 24 at 10:52

rtybasertybase

11.6k31534

2

$begingroup$

See the difference between this one and
$frac{x+2^2x+3^2x+...+n^2 x}{n^3}$ is $frac{{x}+{2^2x}+{3^2x}+...+{n^2 x}}{n^3}$, which is smaller than $frac{n^2}{n^3}=frac 1n to 0$; here ${y}=y-[y]$. So your expression has the same limit $frac 13$ as $frac{x+2^2x+3^2x+...+n^2 x}{n^3}$.

share|cite|improve this answer

answered Mar 24 at 10:41

Yu DingYu Ding

7287

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Probably a typo: makes more sense to say it's less than $frac{n}{n^3}$ instead of $frac{n^2}{n^3}$ as there are $n$ terms in the numerator each of which is less than $1$.
  $endgroup$
  – Winther
  Mar 24 at 15:04
  

  
  
  
  

add a comment | 

2

$begingroup$

See the difference between this one and
$frac{x+2^2x+3^2x+...+n^2 x}{n^3}$ is $frac{{x}+{2^2x}+{3^2x}+...+{n^2 x}}{n^3}$, which is smaller than $frac{n^2}{n^3}=frac 1n to 0$; here ${y}=y-[y]$. So your expression has the same limit $frac 13$ as $frac{x+2^2x+3^2x+...+n^2 x}{n^3}$.

share|cite|improve this answer

answered Mar 24 at 10:41

Yu DingYu Ding

7287

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Probably a typo: makes more sense to say it's less than $frac{n}{n^3}$ instead of $frac{n^2}{n^3}$ as there are $n$ terms in the numerator each of which is less than $1$.
  $endgroup$
  – Winther
  Mar 24 at 15:04
  

  
  
  
  

add a comment | 

$begingroup$

See the difference between this one and
$frac{x+2^2x+3^2x+...+n^2 x}{n^3}$ is $frac{{x}+{2^2x}+{3^2x}+...+{n^2 x}}{n^3}$, which is smaller than $frac{n^2}{n^3}=frac 1n to 0$; here ${y}=y-[y]$. So your expression has the same limit $frac 13$ as $frac{x+2^2x+3^2x+...+n^2 x}{n^3}$.

share|cite|improve this answer

answered Mar 24 at 10:41

Yu DingYu Ding

7287

$endgroup$

share|cite|improve this answer

answered Mar 24 at 10:41

Yu DingYu Ding

7287

answered Mar 24 at 10:41

Yu DingYu Ding

7287

answered Mar 24 at 10:41

Yu DingYu Ding

7287