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Order the following expressions by their size

Evaluation of the sum $sum_k = 0^lfloor a/b rfloor left lfloor fraca - kbc right rfloor$How to prove $left lceil fracnm right rceil = left lfloor fracn+m-1m right rfloor$?Simple problem on restricted partitionFind the chance that $a^3 + b^3 equiv 0 (mod 3)$Dealing with floor function in binomial coefficientsIdentities involving binomial coefficients, floors, and ceilingsAbout Enflo's paper.Evaluating the “limit function” $f_n$ with $n to infty$The least upper bound for $fracsum_i=0^mbinommix^isum_i=0^mx^i$Formula for partial sum of binoms: $sum_k=0^lfloor n/2 rfloor binom nk binom mk$

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Order the following expressions by their size (suppose n is very large):

$$binom2nn-1, binom2nn,binom2n10, n!, n^sqrtn,(sqrtn)^sqrtn, n^15, n^logn, (log n)^n, log (n^n), 2^n $$

I know that $left(beginmatrix 2n \ 10endmatrixright) leleft(beginmatrix 2n \ n-1endmatrixright) le left(beginmatrix 2n \ nendmatrixright)$ because the combination number $left(beginmatrix n \ kendmatrixright)$ increases the closer it gets to $left(beginmatrix n \ lceilfracn2rceilendmatrixright) = left(beginmatrix n \ lfloorfracn2rfloorendmatrixright)$ I also know that $ n^15 le n! le n^n$

combinatorics

share|cite|improve this question

edited Mar 18 at 12:48

awkward

6,71511025

asked Mar 17 at 23:15

J. LastinJ. Lastin

1439

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Relevant link for relating the binomial coefficients to the others: central binomial coefficient.
  $endgroup$
  – JMoravitz
  Mar 17 at 23:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  In General: Logs $ll$ Polynomials $ll$ exponentials $ll$ factorials $ll n^n$. It will also help to remember that $x = e^ln x$ for $x > 0$.
  $endgroup$
  – JavaMan
  Mar 18 at 2:23
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Use the Stirling formula to show that $binom2nn$ is on the order of $frac4^nsqrtn$.
  $endgroup$
  – Alexander Burstein
  Mar 18 at 4:43
  

  
  
  
  

add a comment | 

0

0

$begingroup$

Order the following expressions by their size (suppose n is very large):

$$binom2nn-1, binom2nn,binom2n10, n!, n^sqrtn,(sqrtn)^sqrtn, n^15, n^logn, (log n)^n, log (n^n), 2^n $$

I know that $left(beginmatrix 2n \ 10endmatrixright) leleft(beginmatrix 2n \ n-1endmatrixright) le left(beginmatrix 2n \ nendmatrixright)$ because the combination number $left(beginmatrix n \ kendmatrixright)$ increases the closer it gets to $left(beginmatrix n \ lceilfracn2rceilendmatrixright) = left(beginmatrix n \ lfloorfracn2rfloorendmatrixright)$ I also know that $ n^15 le n! le n^n$

combinatorics

share|cite|improve this question

edited Mar 18 at 12:48

awkward

6,71511025

asked Mar 17 at 23:15

J. LastinJ. Lastin

1439

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Relevant link for relating the binomial coefficients to the others: central binomial coefficient.
  $endgroup$
  – JMoravitz
  Mar 17 at 23:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  In General: Logs $ll$ Polynomials $ll$ exponentials $ll$ factorials $ll n^n$. It will also help to remember that $x = e^ln x$ for $x > 0$.
  $endgroup$
  – JavaMan
  Mar 18 at 2:23
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Use the Stirling formula to show that $binom2nn$ is on the order of $frac4^nsqrtn$.
  $endgroup$
  – Alexander Burstein
  Mar 18 at 4:43
  

  
  
  
  

add a comment | 

$begingroup$

Order the following expressions by their size (suppose n is very large):

$$binom2nn-1, binom2nn,binom2n10, n!, n^sqrtn,(sqrtn)^sqrtn, n^15, n^logn, (log n)^n, log (n^n), 2^n $$

I know that $left(beginmatrix 2n \ 10endmatrixright) leleft(beginmatrix 2n \ n-1endmatrixright) le left(beginmatrix 2n \ nendmatrixright)$ because the combination number $left(beginmatrix n \ kendmatrixright)$ increases the closer it gets to $left(beginmatrix n \ lceilfracn2rceilendmatrixright) = left(beginmatrix n \ lfloorfracn2rfloorendmatrixright)$ I also know that $ n^15 le n! le n^n$

combinatorics

share|cite|improve this question

edited Mar 18 at 12:48

awkward

6,71511025

asked Mar 17 at 23:15

J. LastinJ. Lastin

1439

$endgroup$

share|cite|improve this question

edited Mar 18 at 12:48

awkward

6,71511025

asked Mar 17 at 23:15

J. LastinJ. Lastin

1439

share|cite|improve this question

edited Mar 18 at 12:48

awkward

6,71511025

asked Mar 17 at 23:15

J. LastinJ. Lastin

1439

edited Mar 18 at 12:48

awkward

6,71511025

edited Mar 18 at 12:48

awkward

6,71511025

asked Mar 17 at 23:15

J. LastinJ. Lastin

1439

asked Mar 17 at 23:15

J. LastinJ. Lastin

1439