Find analytic solution: for what $n_0$, $forall n > n_0$ : $n cdot p ^n-1 leq delta$ holds?

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Suppose $p$ is a fixed number in $(0, 1)$ and $delta$ is a small positive number s.t. $ 0 < delta < p$.

What is $n_0$ such that for any $n > n_0$, the following holds:
$$
n cdot p ^n-1 leq delta
$$

real-analysis inequality approximation

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asked Jul 26 at 13:41

Daniel

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up vote
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down vote

favorite

Suppose $p$ is a fixed number in $(0, 1)$ and $delta$ is a small positive number s.t. $ 0 < delta < p$.

What is $n_0$ such that for any $n > n_0$, the following holds:
$$
n cdot p ^n-1 leq delta
$$

real-analysis inequality approximation

share|cite|improve this question

asked Jul 26 at 13:41

Daniel

1,059921

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Suppose $p$ is a fixed number in $(0, 1)$ and $delta$ is a small positive number s.t. $ 0 < delta < p$.

What is $n_0$ such that for any $n > n_0$, the following holds:
$$
n cdot p ^n-1 leq delta
$$

real-analysis inequality approximation

share|cite|improve this question

asked Jul 26 at 13:41

Daniel

1,059921

Suppose $p$ is a fixed number in $(0, 1)$ and $delta$ is a small positive number s.t. $ 0 < delta < p$.

What is $n_0$ such that for any $n > n_0$, the following holds:
$$
n cdot p ^n-1 leq delta
$$

real-analysis inequality approximation

share|cite|improve this question

asked Jul 26 at 13:41

Daniel

1,059921

share|cite|improve this question

share|cite|improve this question

asked Jul 26 at 13:41

Daniel

1,059921

asked Jul 26 at 13:41

Daniel

1,059921

asked Jul 26 at 13:41

Daniel

1,059921

1,059921

add a comment | 

add a comment | 

3 Answers
3

active

oldest

votes

up vote
2
down vote



accepted

Set $p:= e^-y$, where $0 <y$, real

Now consider:

$n e^-y(n-1)=dfracne^y(n-1) =$

$dfracn1+ y(n-1)+ y^2(n-1)^2/2 +....$

$ lt dfrac2ny^2(n-1)^2 lt$

$(1/y^2)dfrac2n(n/2)^2= (8/y^2)(1/n).$

Let $epsilon >0$ be given.

Archimedean principle:

There is a $n_0$, positive integer with

$n_0 > (1/epsilon)(8/y^2).$

For $nge n_0$ we have

$ne^-y(n-1) lt (8/y^2)(1/n)le (8/y^2)(1/n_0) lt epsilon.$

Used: $(n-1)^2 >(n/2)^2$ , $n ge 3.$

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edited Jul 27 at 7:07

answered Jul 26 at 14:25

Peter Szilas

7,8702617

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up vote
1
down vote

Let $r=frac 1p-1.$ That is, $p=frac 11+r.$ We have $r>0.$

If $ngeq 4$ then by the Binomial Theorem we have $$frac 1p^n-1=(1+r)^n-1=1+r(n-1)+r^2frac (n-1)(n-2)2+...>$$ $$>r^2frac (n-1)(n-2)2$$ so we have $$frac 1np^n-1>r^2 frac (n-1)(n-2)2n=frac r^22(n-3+2/n)>frac r^2 2( n-3)$$ so we have $np^n-1< frac 2r^2(n-3) .$ And the rest is obvious.

I felt like giving an answer using only the most elementary means (not even Bernoulli's Inequality).

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answered Jul 26 at 18:53

DanielWainfleet

31.5k31542

• 
  
  
  

  
  

  
  
  
  
  
  

  
  +1 for the simple solution.
  – Paramanand Singh
  Jul 27 at 3:49
  

  
  
  
  

add a comment | 

up vote
0
down vote

$p,qin(0,1)$, $p=1-q$, (Pick $qin(0,1)$ so that this equality holds).
We get $delta geq p^n-1n=(1-q)^n-1ngeq(1-(n-1)q)n $ (using Bernouli`s inequality).
$delta geq n-n^2q-qn leftrightarrow n^2q+(q-1)n+deltageq0$. From here you can find the roots of the polynomial (Relative to $n$), take the bigger one ($n_0$), and since this is a parabola, the inequality holds for all $ngeq n_0$

share|cite|improve this answer

answered Jul 26 at 14:52

Sar

40410

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

Set $p:= e^-y$, where $0 <y$, real

Now consider:

$n e^-y(n-1)=dfracne^y(n-1) =$

$dfracn1+ y(n-1)+ y^2(n-1)^2/2 +....$

$ lt dfrac2ny^2(n-1)^2 lt$

$(1/y^2)dfrac2n(n/2)^2= (8/y^2)(1/n).$

Let $epsilon >0$ be given.

Archimedean principle:

There is a $n_0$, positive integer with

$n_0 > (1/epsilon)(8/y^2).$

For $nge n_0$ we have

$ne^-y(n-1) lt (8/y^2)(1/n)le (8/y^2)(1/n_0) lt epsilon.$

Used: $(n-1)^2 >(n/2)^2$ , $n ge 3.$

share|cite|improve this answer

edited Jul 27 at 7:07

answered Jul 26 at 14:25

Peter Szilas

7,8702617

add a comment | 

up vote
2
down vote



accepted

Set $p:= e^-y$, where $0 <y$, real

Now consider:

$n e^-y(n-1)=dfracne^y(n-1) =$

$dfracn1+ y(n-1)+ y^2(n-1)^2/2 +....$

$ lt dfrac2ny^2(n-1)^2 lt$

$(1/y^2)dfrac2n(n/2)^2= (8/y^2)(1/n).$

Let $epsilon >0$ be given.

Archimedean principle:

There is a $n_0$, positive integer with

$n_0 > (1/epsilon)(8/y^2).$

For $nge n_0$ we have

$ne^-y(n-1) lt (8/y^2)(1/n)le (8/y^2)(1/n_0) lt epsilon.$

Used: $(n-1)^2 >(n/2)^2$ , $n ge 3.$

share|cite|improve this answer

edited Jul 27 at 7:07

answered Jul 26 at 14:25

Peter Szilas

7,8702617

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

Set $p:= e^-y$, where $0 <y$, real

Now consider:

$n e^-y(n-1)=dfracne^y(n-1) =$

$dfracn1+ y(n-1)+ y^2(n-1)^2/2 +....$

$ lt dfrac2ny^2(n-1)^2 lt$

$(1/y^2)dfrac2n(n/2)^2= (8/y^2)(1/n).$

Let $epsilon >0$ be given.

Archimedean principle:

There is a $n_0$, positive integer with

$n_0 > (1/epsilon)(8/y^2).$

For $nge n_0$ we have

$ne^-y(n-1) lt (8/y^2)(1/n)le (8/y^2)(1/n_0) lt epsilon.$

Used: $(n-1)^2 >(n/2)^2$ , $n ge 3.$

share|cite|improve this answer

edited Jul 27 at 7:07

answered Jul 26 at 14:25

Peter Szilas

7,8702617

Set $p:= e^-y$, where $0 <y$, real

Now consider:

$n e^-y(n-1)=dfracne^y(n-1) =$

$dfracn1+ y(n-1)+ y^2(n-1)^2/2 +....$

$ lt dfrac2ny^2(n-1)^2 lt$

$(1/y^2)dfrac2n(n/2)^2= (8/y^2)(1/n).$

Let $epsilon >0$ be given.

Archimedean principle:

There is a $n_0$, positive integer with

$n_0 > (1/epsilon)(8/y^2).$

For $nge n_0$ we have

$ne^-y(n-1) lt (8/y^2)(1/n)le (8/y^2)(1/n_0) lt epsilon.$

Used: $(n-1)^2 >(n/2)^2$ , $n ge 3.$

share|cite|improve this answer

edited Jul 27 at 7:07

answered Jul 26 at 14:25

Peter Szilas

7,8702617

share|cite|improve this answer

share|cite|improve this answer

edited Jul 27 at 7:07

edited Jul 27 at 7:07

edited Jul 27 at 7:07

answered Jul 26 at 14:25

Peter Szilas

7,8702617

answered Jul 26 at 14:25

Peter Szilas

7,8702617

answered Jul 26 at 14:25

Peter Szilas

7,8702617

7,8702617

add a comment | 

add a comment | 

up vote
1
down vote

Let $r=frac 1p-1.$ That is, $p=frac 11+r.$ We have $r>0.$

If $ngeq 4$ then by the Binomial Theorem we have $$frac 1p^n-1=(1+r)^n-1=1+r(n-1)+r^2frac (n-1)(n-2)2+...>$$ $$>r^2frac (n-1)(n-2)2$$ so we have $$frac 1np^n-1>r^2 frac (n-1)(n-2)2n=frac r^22(n-3+2/n)>frac r^2 2( n-3)$$ so we have $np^n-1< frac 2r^2(n-3) .$ And the rest is obvious.

I felt like giving an answer using only the most elementary means (not even Bernoulli's Inequality).

share|cite|improve this answer

answered Jul 26 at 18:53

DanielWainfleet

31.5k31542

• 
  
  
  

  
  

  
  
  
  
  
  

  
  +1 for the simple solution.
  – Paramanand Singh
  Jul 27 at 3:49
  

  
  
  
  

add a comment | 

up vote
1
down vote

Let $r=frac 1p-1.$ That is, $p=frac 11+r.$ We have $r>0.$

If $ngeq 4$ then by the Binomial Theorem we have $$frac 1p^n-1=(1+r)^n-1=1+r(n-1)+r^2frac (n-1)(n-2)2+...>$$ $$>r^2frac (n-1)(n-2)2$$ so we have $$frac 1np^n-1>r^2 frac (n-1)(n-2)2n=frac r^22(n-3+2/n)>frac r^2 2( n-3)$$ so we have $np^n-1< frac 2r^2(n-3) .$ And the rest is obvious.

I felt like giving an answer using only the most elementary means (not even Bernoulli's Inequality).

share|cite|improve this answer

answered Jul 26 at 18:53

DanielWainfleet

31.5k31542

• 
  
  
  

  
  

  
  
  
  
  
  

  
  +1 for the simple solution.
  – Paramanand Singh
  Jul 27 at 3:49
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

Let $r=frac 1p-1.$ That is, $p=frac 11+r.$ We have $r>0.$

If $ngeq 4$ then by the Binomial Theorem we have $$frac 1p^n-1=(1+r)^n-1=1+r(n-1)+r^2frac (n-1)(n-2)2+...>$$ $$>r^2frac (n-1)(n-2)2$$ so we have $$frac 1np^n-1>r^2 frac (n-1)(n-2)2n=frac r^22(n-3+2/n)>frac r^2 2( n-3)$$ so we have $np^n-1< frac 2r^2(n-3) .$ And the rest is obvious.

I felt like giving an answer using only the most elementary means (not even Bernoulli's Inequality).

share|cite|improve this answer

answered Jul 26 at 18:53

DanielWainfleet

31.5k31542

Let $r=frac 1p-1.$ That is, $p=frac 11+r.$ We have $r>0.$

If $ngeq 4$ then by the Binomial Theorem we have $$frac 1p^n-1=(1+r)^n-1=1+r(n-1)+r^2frac (n-1)(n-2)2+...>$$ $$>r^2frac (n-1)(n-2)2$$ so we have $$frac 1np^n-1>r^2 frac (n-1)(n-2)2n=frac r^22(n-3+2/n)>frac r^2 2( n-3)$$ so we have $np^n-1< frac 2r^2(n-3) .$ And the rest is obvious.

I felt like giving an answer using only the most elementary means (not even Bernoulli's Inequality).

share|cite|improve this answer

answered Jul 26 at 18:53

DanielWainfleet

31.5k31542

share|cite|improve this answer

share|cite|improve this answer

answered Jul 26 at 18:53

DanielWainfleet

31.5k31542

answered Jul 26 at 18:53

DanielWainfleet

31.5k31542

answered Jul 26 at 18:53

DanielWainfleet

31.5k31542

31.5k31542

• 
  
  
  

  
  

  
  
  
  
  
  

  
  +1 for the simple solution.
  – Paramanand Singh
  Jul 27 at 3:49
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  +1 for the simple solution.
  – Paramanand Singh
  Jul 27 at 3:49
  

  
  
  
  

+1 for the simple solution.
– Paramanand Singh
Jul 27 at 3:49

+1 for the simple solution.
– Paramanand Singh
Jul 27 at 3:49

add a comment | 

up vote
0
down vote

$p,qin(0,1)$, $p=1-q$, (Pick $qin(0,1)$ so that this equality holds).
We get $delta geq p^n-1n=(1-q)^n-1ngeq(1-(n-1)q)n $ (using Bernouli`s inequality).
$delta geq n-n^2q-qn leftrightarrow n^2q+(q-1)n+deltageq0$. From here you can find the roots of the polynomial (Relative to $n$), take the bigger one ($n_0$), and since this is a parabola, the inequality holds for all $ngeq n_0$

share|cite|improve this answer

answered Jul 26 at 14:52

Sar

40410

add a comment | 

up vote
0
down vote

$p,qin(0,1)$, $p=1-q$, (Pick $qin(0,1)$ so that this equality holds).
We get $delta geq p^n-1n=(1-q)^n-1ngeq(1-(n-1)q)n $ (using Bernouli`s inequality).
$delta geq n-n^2q-qn leftrightarrow n^2q+(q-1)n+deltageq0$. From here you can find the roots of the polynomial (Relative to $n$), take the bigger one ($n_0$), and since this is a parabola, the inequality holds for all $ngeq n_0$

share|cite|improve this answer

answered Jul 26 at 14:52

Sar

40410

add a comment | 

up vote
0
down vote

up vote
0
down vote

$p,qin(0,1)$, $p=1-q$, (Pick $qin(0,1)$ so that this equality holds).
We get $delta geq p^n-1n=(1-q)^n-1ngeq(1-(n-1)q)n $ (using Bernouli`s inequality).
$delta geq n-n^2q-qn leftrightarrow n^2q+(q-1)n+deltageq0$. From here you can find the roots of the polynomial (Relative to $n$), take the bigger one ($n_0$), and since this is a parabola, the inequality holds for all $ngeq n_0$

share|cite|improve this answer

answered Jul 26 at 14:52

Sar

40410

$p,qin(0,1)$, $p=1-q$, (Pick $qin(0,1)$ so that this equality holds).
We get $delta geq p^n-1n=(1-q)^n-1ngeq(1-(n-1)q)n $ (using Bernouli`s inequality).
$delta geq n-n^2q-qn leftrightarrow n^2q+(q-1)n+deltageq0$. From here you can find the roots of the polynomial (Relative to $n$), take the bigger one ($n_0$), and since this is a parabola, the inequality holds for all $ngeq n_0$

share|cite|improve this answer

answered Jul 26 at 14:52

Sar

40410

share|cite|improve this answer

share|cite|improve this answer

answered Jul 26 at 14:52

Sar

40410

answered Jul 26 at 14:52

Sar

40410

answered Jul 26 at 14:52

Sar

40410

40410

add a comment | 

add a comment | 

 

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