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How do i show$lim_nrightarrow infty frac[(n+1)(n+2)…(2n)]^frac1nn=frac4e$ without using integration?

Proving that $lim_n rightarrowinfty int_0^fracpi2 sin(t^n) dt =0$Prove that $lim_xtoinfty frace^xx^n=infty$ without using L'hôpital's ruleHow to show that $lim_nrightarrow infty (1+fracxn)^n=e^x$$lim_nrightarrowinftyfrac1nsum_k=1^nfrackk^2+1$Show that $lim_ntoinftyfrac2^nn^ln(n)=infty$Find the limit $lim_limitsxto 0^+left( e^frac1sin x-e^frac1xright)$Show that $lim_t rightarrow infty tF(t) = 0.$Show that $lim_x rightarrow infty f(x)$ exists by the given condition.How can I evaluate $lim_n rightarrow infty int_n^infty fracn^2 arctan frac1xx^2+n^2 dx$?Why is $lim_x rightarrow -inftyfracxsqrtx^2 = -1$ and not $1$?

1

2

$begingroup$

Without using integration show that $$lim_nrightarrow infty frac[(n+1)(n+2)...(2n)]^frac1nn=frac4e$$. It could have been easier with integration, but I cannot proceed with this one. Can anyone please give me any hint or the solution... I will be very grateful. Please.

real-analysis integration limits

share|cite|improve this question

edited Apr 30 at 12:01

YuiTo Cheng

3,21161145

asked Apr 30 at 6:47

user655791

$endgroup$

add a comment | 

1

2

$begingroup$

Without using integration show that $$lim_nrightarrow infty frac[(n+1)(n+2)...(2n)]^frac1nn=frac4e$$. It could have been easier with integration, but I cannot proceed with this one. Can anyone please give me any hint or the solution... I will be very grateful. Please.

real-analysis integration limits

share|cite|improve this question

edited Apr 30 at 12:01

YuiTo Cheng

3,21161145

asked Apr 30 at 6:47

user655791

$endgroup$

add a comment | 

$begingroup$

Without using integration show that $$lim_nrightarrow infty frac[(n+1)(n+2)...(2n)]^frac1nn=frac4e$$. It could have been easier with integration, but I cannot proceed with this one. Can anyone please give me any hint or the solution... I will be very grateful. Please.

real-analysis integration limits

share|cite|improve this question

edited Apr 30 at 12:01

YuiTo Cheng

3,21161145

asked Apr 30 at 6:47

user655791

$endgroup$

share|cite|improve this question

edited Apr 30 at 12:01

YuiTo Cheng

3,21161145

asked Apr 30 at 6:47

user655791

share|cite|improve this question

edited Apr 30 at 12:01

YuiTo Cheng

3,21161145

asked Apr 30 at 6:47

user655791

edited Apr 30 at 12:01

YuiTo Cheng

3,21161145

edited Apr 30 at 12:01

YuiTo Cheng

3,21161145

3 Answers
3

active

oldest

votes

5

$begingroup$

By Stirling,

$$frac1nsqrt[n]frac(2n)!n!simfrac1nfracsqrt[2n]4pi nleft(dfrac2neright)^2sqrt[n]4pi ndfrac netofrac 4e.$$

share|cite|improve this answer

answered Apr 30 at 7:23

Yves DaoustYves Daoust

136k676237

$endgroup$

add a comment | 

3

$begingroup$

If we let $a_n=frac[(n+1)(n+2)...(2n)]n^n=frac(2n)!n!n^n$, then $frac[(n+1)(n+2)...(2n)]^frac1nn=(a_n)^frac1n$. Then $fraca_n+1a_n=frac(2n+2)!(2n)!×fracn!(n+1)!×fracn^n(n+1)^n+1=frac(2n+1)(2n+2)(n+1)×fracn^n(n+1)^n+1=frac2(2n+1)(n+1)×fracn^n(n+1)^n+1=frac4n(1+frac12n)n(1+frac1n)×(1+frac1n)^-nrightarrow4×e^-1$.

Edit:explaining the last step explicitly... note that as $nrightarrow infty, frac1n,frac12nrightarrow 0$, so we can replace $frac12n$ with $frac1n$. Now hence, $frac4n(1+frac12n)n(1+frac1n)×(1+frac1n)^-n=4×(1+frac1n)^-n rightarrow 4× e^-1$

share|cite|improve this answer

edited Apr 30 at 7:32

answered Apr 30 at 7:01

Shamim AkhtarShamim Akhtar

54919

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The last step is a little too fast. Can you justify why you can replace the two factors by their limits $4$ and $e^-1$, while you still have an infinite product for all $n$.
  $endgroup$
  – Yves Daoust
  Apr 30 at 7:14
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I justified....
  $endgroup$
  – Shamim Akhtar
  Apr 30 at 7:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  No, you didn't add a word. How is it that $sqrt[n]prod_1,n(1+frac1k)^kto e$ ?
  $endgroup$
  – Yves Daoust
  Apr 30 at 7:34
  
  

  
  
  
  

add a comment | 

-1

$begingroup$

By taking logs of both sides we can get:
$$L=lim_ntoinftyfracleft[(n+1)(n+2)(n+3)...(2n)right]^1/nn$$
$$ln(L)=lim_ntoinftyfrac1nlnleft[(n+1)(n+2)(n+3)...(2n)right]-ln(n)$$
$$ln(L)=lim_ntoinftyfrac1nsum_i=1^nln(n+i)-ln(n)$$
And I am sure there is a formula for the summation of logs in this form. Alternatively we can say:
$$ln(L)=lim_ntoinftyfrac1nlnleft(frac(2n)!n!right)-ln(n)$$

share|cite|improve this answer

answered Apr 30 at 15:14

Henry LeeHenry Lee

2,288319

$endgroup$

add a comment | 

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5

$begingroup$

By Stirling,

$$frac1nsqrt[n]frac(2n)!n!simfrac1nfracsqrt[2n]4pi nleft(dfrac2neright)^2sqrt[n]4pi ndfrac netofrac 4e.$$

share|cite|improve this answer

answered Apr 30 at 7:23

Yves DaoustYves Daoust

136k676237

$endgroup$

add a comment | 

5

$begingroup$

By Stirling,

$$frac1nsqrt[n]frac(2n)!n!simfrac1nfracsqrt[2n]4pi nleft(dfrac2neright)^2sqrt[n]4pi ndfrac netofrac 4e.$$

share|cite|improve this answer

answered Apr 30 at 7:23

Yves DaoustYves Daoust

136k676237

$endgroup$

add a comment | 

$begingroup$

By Stirling,

$$frac1nsqrt[n]frac(2n)!n!simfrac1nfracsqrt[2n]4pi nleft(dfrac2neright)^2sqrt[n]4pi ndfrac netofrac 4e.$$

share|cite|improve this answer

answered Apr 30 at 7:23

Yves DaoustYves Daoust

136k676237

$endgroup$

share|cite|improve this answer

answered Apr 30 at 7:23

Yves DaoustYves Daoust

136k676237

answered Apr 30 at 7:23

Yves DaoustYves Daoust

136k676237

answered Apr 30 at 7:23

Yves DaoustYves Daoust

136k676237

3

$begingroup$

If we let $a_n=frac[(n+1)(n+2)...(2n)]n^n=frac(2n)!n!n^n$, then $frac[(n+1)(n+2)...(2n)]^frac1nn=(a_n)^frac1n$. Then $fraca_n+1a_n=frac(2n+2)!(2n)!×fracn!(n+1)!×fracn^n(n+1)^n+1=frac(2n+1)(2n+2)(n+1)×fracn^n(n+1)^n+1=frac2(2n+1)(n+1)×fracn^n(n+1)^n+1=frac4n(1+frac12n)n(1+frac1n)×(1+frac1n)^-nrightarrow4×e^-1$.

Edit:explaining the last step explicitly... note that as $nrightarrow infty, frac1n,frac12nrightarrow 0$, so we can replace $frac12n$ with $frac1n$. Now hence, $frac4n(1+frac12n)n(1+frac1n)×(1+frac1n)^-n=4×(1+frac1n)^-n rightarrow 4× e^-1$

share|cite|improve this answer

edited Apr 30 at 7:32

answered Apr 30 at 7:01

Shamim AkhtarShamim Akhtar

54919

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The last step is a little too fast. Can you justify why you can replace the two factors by their limits $4$ and $e^-1$, while you still have an infinite product for all $n$.
  $endgroup$
  – Yves Daoust
  Apr 30 at 7:14
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I justified....
  $endgroup$
  – Shamim Akhtar
  Apr 30 at 7:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  No, you didn't add a word. How is it that $sqrt[n]prod_1,n(1+frac1k)^kto e$ ?
  $endgroup$
  – Yves Daoust
  Apr 30 at 7:34
  
  

  
  
  
  

add a comment | 

3

$begingroup$

If we let $a_n=frac[(n+1)(n+2)...(2n)]n^n=frac(2n)!n!n^n$, then $frac[(n+1)(n+2)...(2n)]^frac1nn=(a_n)^frac1n$. Then $fraca_n+1a_n=frac(2n+2)!(2n)!×fracn!(n+1)!×fracn^n(n+1)^n+1=frac(2n+1)(2n+2)(n+1)×fracn^n(n+1)^n+1=frac2(2n+1)(n+1)×fracn^n(n+1)^n+1=frac4n(1+frac12n)n(1+frac1n)×(1+frac1n)^-nrightarrow4×e^-1$.

Edit:explaining the last step explicitly... note that as $nrightarrow infty, frac1n,frac12nrightarrow 0$, so we can replace $frac12n$ with $frac1n$. Now hence, $frac4n(1+frac12n)n(1+frac1n)×(1+frac1n)^-n=4×(1+frac1n)^-n rightarrow 4× e^-1$

share|cite|improve this answer

edited Apr 30 at 7:32

answered Apr 30 at 7:01

Shamim AkhtarShamim Akhtar

54919

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The last step is a little too fast. Can you justify why you can replace the two factors by their limits $4$ and $e^-1$, while you still have an infinite product for all $n$.
  $endgroup$
  – Yves Daoust
  Apr 30 at 7:14
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I justified....
  $endgroup$
  – Shamim Akhtar
  Apr 30 at 7:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  No, you didn't add a word. How is it that $sqrt[n]prod_1,n(1+frac1k)^kto e$ ?
  $endgroup$
  – Yves Daoust
  Apr 30 at 7:34
  
  

  
  
  
  

add a comment | 

$begingroup$

If we let $a_n=frac[(n+1)(n+2)...(2n)]n^n=frac(2n)!n!n^n$, then $frac[(n+1)(n+2)...(2n)]^frac1nn=(a_n)^frac1n$. Then $fraca_n+1a_n=frac(2n+2)!(2n)!×fracn!(n+1)!×fracn^n(n+1)^n+1=frac(2n+1)(2n+2)(n+1)×fracn^n(n+1)^n+1=frac2(2n+1)(n+1)×fracn^n(n+1)^n+1=frac4n(1+frac12n)n(1+frac1n)×(1+frac1n)^-nrightarrow4×e^-1$.

Edit:explaining the last step explicitly... note that as $nrightarrow infty, frac1n,frac12nrightarrow 0$, so we can replace $frac12n$ with $frac1n$. Now hence, $frac4n(1+frac12n)n(1+frac1n)×(1+frac1n)^-n=4×(1+frac1n)^-n rightarrow 4× e^-1$

share|cite|improve this answer

edited Apr 30 at 7:32

answered Apr 30 at 7:01

Shamim AkhtarShamim Akhtar

54919

$endgroup$

share|cite|improve this answer

edited Apr 30 at 7:32

answered Apr 30 at 7:01

Shamim AkhtarShamim Akhtar

54919

answered Apr 30 at 7:01

Shamim AkhtarShamim Akhtar

54919

answered Apr 30 at 7:01

Shamim AkhtarShamim Akhtar

54919

-1

$begingroup$

By taking logs of both sides we can get:
$$L=lim_ntoinftyfracleft[(n+1)(n+2)(n+3)...(2n)right]^1/nn$$
$$ln(L)=lim_ntoinftyfrac1nlnleft[(n+1)(n+2)(n+3)...(2n)right]-ln(n)$$
$$ln(L)=lim_ntoinftyfrac1nsum_i=1^nln(n+i)-ln(n)$$
And I am sure there is a formula for the summation of logs in this form. Alternatively we can say:
$$ln(L)=lim_ntoinftyfrac1nlnleft(frac(2n)!n!right)-ln(n)$$

share|cite|improve this answer

answered Apr 30 at 15:14

Henry LeeHenry Lee

2,288319

$endgroup$

add a comment | 

-1

$begingroup$

By taking logs of both sides we can get:
$$L=lim_ntoinftyfracleft[(n+1)(n+2)(n+3)...(2n)right]^1/nn$$
$$ln(L)=lim_ntoinftyfrac1nlnleft[(n+1)(n+2)(n+3)...(2n)right]-ln(n)$$
$$ln(L)=lim_ntoinftyfrac1nsum_i=1^nln(n+i)-ln(n)$$
And I am sure there is a formula for the summation of logs in this form. Alternatively we can say:
$$ln(L)=lim_ntoinftyfrac1nlnleft(frac(2n)!n!right)-ln(n)$$

share|cite|improve this answer

answered Apr 30 at 15:14

Henry LeeHenry Lee

2,288319

$endgroup$

add a comment | 

$begingroup$

By taking logs of both sides we can get:
$$L=lim_ntoinftyfracleft[(n+1)(n+2)(n+3)...(2n)right]^1/nn$$
$$ln(L)=lim_ntoinftyfrac1nlnleft[(n+1)(n+2)(n+3)...(2n)right]-ln(n)$$
$$ln(L)=lim_ntoinftyfrac1nsum_i=1^nln(n+i)-ln(n)$$
And I am sure there is a formula for the summation of logs in this form. Alternatively we can say:
$$ln(L)=lim_ntoinftyfrac1nlnleft(frac(2n)!n!right)-ln(n)$$

share|cite|improve this answer

answered Apr 30 at 15:14

Henry LeeHenry Lee

2,288319

$endgroup$

share|cite|improve this answer

answered Apr 30 at 15:14

Henry LeeHenry Lee

2,288319

answered Apr 30 at 15:14

Henry LeeHenry Lee

2,288319

answered Apr 30 at 15:14

Henry LeeHenry Lee

2,288319