Conference rusure::math

    >         lim (n**b)/(a**n)  
    >        n->inf
    
     we can try this:
    
        n=oo
        ---
        \        n^b/a^n
        /
        ---
        n=1
    
    now, if the above is convergent series, then the nth term, as n->oo
    must get smaller and smaller, and will tend to zero.
    so, all what we need is to prove that the above series converges.
    try the ratio test, take ratio of n+1 term to n term, and take n->oo
    
                 (n+1)^b/a^(n+1)      (n+1)^b
                 ---------------- = ---------- = 1/a  as n->oo
                     n^b/a^n           a n^b
    
    but a>1, so 1/a < 1 , so the series converges , so lim n->oo n^b/a^n=0
    
    

    >        lim (n!)/(n*log(n)) = 0
    >        n->inf
    
    use ratio test
    
                  (n+1)! /( (n+1)*log(n+1) )     (n+1) n! n*log(n)
                  -------------------------- = --------------------
                      n!/n*log(n)                n! (n+1)*log(n+1)
    
        n * log(n)
     = ------------ --> n  as n->oo , i.e. goes to oo.
        log(n+1)
    
    ratio test says the series n!/nlog(n) is convergents, so
    nth term n!/n*log(n) does not go to zero as n->oo, which contredicts 
    what you ask to show ?
    

    Alternate solutions using logs
    
>             lim (n**b)/(a**n)  = 0
>            n->inf

    The log of (n**b)/(a**n) is b log n - n log a.
    a > 1 so log a is positive; n grows much faster than log n, so
    n log a outgrows b log n and so the difference b log n - n log a
    becomes more and more negative as n increases.  Thus (n**b)/(a**n)
    approaches zero as n increases without bound.

>             lim (n**log(n))/[(log(n))**n] = 0
>            n->inf
    
    Again taking logs, log (n**log(n))/[(log(n))**n] is
    (log n)^2 - n log log n, which again becomes more and more
    negative as n increases.  The limit is again zero.
    
>    and      lim (n!)/(n*log(n)) = 0
>            n->inf
    
    There is no way that this limit is zero.  n! grows much faster
    than n log n; the quotient increases without bound as n does.
     
>    where a, b are real constants and a > 1.
    
    Dan

    It could be that the author of .0 is looking for a more basic solution
    than has been provided, one which uses the definition of limit, i.e.,
    given epsilon > 0 there exists N such that, e.g.:
    
    	n > N --> (n^b)/(a^n) < epsilon
    
    	n > N --> (n**log(n))/[(log(n))**n] < epsilon
    
    	~(n > N --> (n!)/(n*log(n)) < epsilon)
    
    and the REAL problem is to find such an N in terms of epsilon.
    
      John

  <<< Note 1667.3 by CSC32::D_DERAMO "Dan D'Eramo, Customer Support Center" >>>

<< ...
   
<< >    and      lim (n!)/(n*log(n)) = 0
<< >            n->inf
    
<<    There is no way that this limit is zero.  n! grows much faster
<<    than n log n; the quotient increases without bound as n does.
     
<< >    where a, b are real constants and a > 1.

This one is tailor-made for Stirling's formula:

	n! = sqrt(2*pi*n)*(n/e)**n*(1+theta/(12*n))

where theta is between 0 and 1 (and, obviously, depends on n).

Based on this, it's quite obvious that the expression in the original note
tends fairly rapidly towards infinity.

My guess is that the original problem should have been

	 lim  (n!)/(n**n) = 0
	n->oo

    ref .2 (me)
    
    >    n * log(n)
    > = ------------ --> n  as n->oo , i.e. goes to oo.
    >    log(n+1)
    > 
    >ratio test says the series n!/nlog(n) is convergents, so
                                           ^^^^^^^^^^^^^^
    >nth term n!/n*log(n) does not go to zero as n->oo, which contredicts 
    >what you ask to show ?
    
    even thought it is clear from context, i wanted to say "is NOT convergent"
    
    /nasser
    

Re .4: I agree, and I worked on such a proof last night but it got too
cumbersome (and late) so I gave up.

Nuggets from my attempt were:

If f(n) = (n^b)/(a^n) then f(2*n) = f(n)* (2^b)/(a^n).

So, if n > N0 = (b+1)*log(2)/log(a) then f(2n) < f(n)/2 and f(n*2^k) < f(n)/2^k

This can be used to find the N for every epsilon < 1, namely

N = N0*2*f(N0)/epsilon

(i.e. start at N0 & go up at least enough octaves to force f(N0*2^k) < epsilon) 

What remains is a simple proof that f(n) is monotonically decreasing for
n > some N1, so that for ALL n > max (N, N1), f(n) < epsilon...

    Thanks for all replies.
    
    Yes, the 3rd one should be related to Stirling's formula.
    
    No, I don't need to find epsilon for the proves.
    
    Oh, I suddenly have so many TA helping me :-)
    
    -Aston