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show that $frac1k^a-1 - frac1(k+1)^a-1 geq frac1zeta(a).k^ac $

1

1

$begingroup$

I want to Apply the acceptance reject method to the zipf distribution.
For that i want to use q(k)= $frac1k^a-1 - frac1(k+1)^a-1$

I have to show there exist c>1, such that $frac1k^a-1 - frac1(k+1)^a-1 geq frac1zeta(a).k^ac $

where $zeta(a)= sum_ngeq1 frac1n^a$ , a>1

real-analysis probability probability-theory probability-distributions simulation

share|cite|improve this question

edited Apr 2 at 17:19

Farouk Deutsch

asked Apr 2 at 16:22

Farouk DeutschFarouk Deutsch

1239

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Use the Taylor expansion of $(1+x)^1-a$ with $x=1/(k+1)$ equivalently $frac1k^a-1 - frac1(k+1)^a-1 = int_k^k+1 (a-1)t^-adt =(a-1)k^-a+(a-1) aint_k^k+1 int_k^t (k^-a-1-u^-a-1)dudt$
  $endgroup$
  – reuns
  Apr 2 at 20:46
  
  

  
  
  
  

add a comment | 

1

1

$begingroup$

I want to Apply the acceptance reject method to the zipf distribution.
For that i want to use q(k)= $frac1k^a-1 - frac1(k+1)^a-1$

I have to show there exist c>1, such that $frac1k^a-1 - frac1(k+1)^a-1 geq frac1zeta(a).k^ac $

where $zeta(a)= sum_ngeq1 frac1n^a$ , a>1

real-analysis probability probability-theory probability-distributions simulation

share|cite|improve this question

edited Apr 2 at 17:19

Farouk Deutsch

asked Apr 2 at 16:22

Farouk DeutschFarouk Deutsch

1239

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Use the Taylor expansion of $(1+x)^1-a$ with $x=1/(k+1)$ equivalently $frac1k^a-1 - frac1(k+1)^a-1 = int_k^k+1 (a-1)t^-adt =(a-1)k^-a+(a-1) aint_k^k+1 int_k^t (k^-a-1-u^-a-1)dudt$
  $endgroup$
  – reuns
  Apr 2 at 20:46
  
  

  
  
  
  

add a comment | 

$begingroup$

I want to Apply the acceptance reject method to the zipf distribution.
For that i want to use q(k)= $frac1k^a-1 - frac1(k+1)^a-1$

I have to show there exist c>1, such that $frac1k^a-1 - frac1(k+1)^a-1 geq frac1zeta(a).k^ac $

where $zeta(a)= sum_ngeq1 frac1n^a$ , a>1

real-analysis probability probability-theory probability-distributions simulation

share|cite|improve this question

edited Apr 2 at 17:19

Farouk Deutsch

asked Apr 2 at 16:22

Farouk DeutschFarouk Deutsch

1239

$endgroup$

share|cite|improve this question

edited Apr 2 at 17:19

Farouk Deutsch

asked Apr 2 at 16:22

Farouk DeutschFarouk Deutsch

1239

share|cite|improve this question

edited Apr 2 at 17:19

Farouk Deutsch

asked Apr 2 at 16:22

Farouk DeutschFarouk Deutsch

1239

asked Apr 2 at 16:22

Farouk DeutschFarouk Deutsch

1239

asked Apr 2 at 16:22

Farouk DeutschFarouk Deutsch

1239