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Limit of an Infinite Series

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Date: 18 Jul 1995 16:38:39 -0400
From: Juan Jose Cukier
Subject: Mathematical Analysis

Dear Dr. Math,

I am a student of computing and while studying for a final
exam of Mathematical Analysis, I came across the following question:
What is the limit of the following expression?

(n!)^(1/n)

As I do know that limit n^(1/n) equals 1, I thought that
(n!)^(1/n) could be expressed as follows:

(n(n-1)(n-1)...1)^(1/n) = n^(1/n) (n-1)^(1/n) ... 1^(1/n)

and the limit would be 1.1.1.1. ... .1 = 1, but I am not sure whether I used
the property properly, since I am working with an infinite number of bases.

Yours faithfully,

-     Juan Jose Cukier     -
<jjc@infool.ba.ar>    Buenos Aires, Argentina.
```

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Date: 19 Jul 1995 09:57:12 -0400
From: Dr. Ethan
Subject: Re: Mathematical Analysis

Well I am afraid that it can't work that way.  Here is another way to look
at it.

n! = n^(n-1) -n^(n-2) etc......

So n^(n-1) is a close upper bound to n!
and

(n^(n-1))^(1/n) = n^((n-1)/n).  This clearly does not go to one as n gets
large.

Similarly for large n n^(n-2) would be a lower bound.  It also does not go
to one for very large n.
. . .

We need to see that we can find a number b greater than one such that
b^n is smaller than n!.  For instance 40! > 13^40 and as n gets bigger we can
choose b bigger and bigger.

So the lim (n!)^(1/n) is bigger than any number you can pick.

For instance, pick 200.  Well I can say that the lim(200^n)^(1/n)
is 200 and once n is bigger than 700 (it actually could probably be
much smaller than 700 but 700 is plenty big) n! is bigger that 200^n so
its limit will be bigger.  Since we can do this for any number we want no
matter how big, the lim (n!)^(1/n) is infinity.

Hope that seems clear.

Ethan Doctor On Call
```

```
Date: 2/29/96 at 17:21:56
From: Doctor Ethan
Subject: Mathematical Analysis

Here is another way to look at this that Leo Broukhis sent in to us.

From Stirling's formula, we know that n! is asymptotic to

n^n * e^-n * sqrt(2*Pi*n)

so (n!)^(1/n) -> n/e

and this clearly approaches infinity as n approaches infinity.

Thanks Leo.

-Doctor Ethan, The Math Forum
```

```
From etrui@zeus.ci.ua.pt
Date: 05/16/96 at 08:18:54
Subject: Suggestion Box

I'd like to suggest adding to your page the information in the 'sci.math'
Newsgroup, which I give you now:

Check this solution:

Problem: Calculate limit of the succession u(n), where:
u(n) = (n!)^(1/n)

1) v(n) = ln[u(n)] = ln(n!)/n. So, u(n) = exp[v(n)].

2) v(n) = Sum[k=1,n, ln(k)/n] = (ln1 + ln2 + ... + ln(n) ) / n
= Sum[k=1,n, ln(k/n)/n] + Sum[k=1,n, ln(n)/n] =
= Sum[k=1,n, ln(k/n)/n] + ln(n)

{Sum[k=1,n, u(n)] stands for the summation of the first 'n' terms of u(n)}

3) Sum[k=1,n, ln(k/n)/n] tends to the integral of ln(x) between 0(+) and
1 (remember definition of Riemann's integral).

4) This integral evaluates as follows:

1
I = S ln(x) dx = [xlnx -x] between 0 and 1.
0
[xlnx-x] = -1 when x=1...

...and tends to 0 when x-> 0(+).

So, I = -1.

5) ln(n) tends to +inf so, from 2, 3 and 4 comes:
v(n) -> +inf.

6) u(n) = exp[v(n)] so, u(n) also tends to +oo  QED

Let's put it more general...

lim u(n) = lim n!^(1/n^p), where 'p' is a positive real number (n->+inf).

v(n)= ln(u(n)) = ln n! / n^p

Taking the same reasoning I used above,

lim v(n) = lim (1/n^[p-1]) [ln n - Sum[k=1,n, ln(k/n)*1/n]

As n->+inf, Sum[...] tends to the integral of ln(x) from 0(+) to 1
(which is -1). So, if...

...p=1 (previous case), then v(n)->+inf, so u(n) ->+inf
...p<1, then v(n)-> +inf, so u(n) -> +inf
...p>1, then v(n)-> 0, so u(n) -> 1     {for ln n/n^a -> 0 when a>0}

Synthetizing...
u(n)-> +inf if p<=1
u(n)-> 1 if p>1

Nice... It looks like the convergence and divergence of the Harmonic
Series  ( 1/n^p ).

Best regards,

rui

----------------------------------
Rui Jorge Ferreira da Costa
Rua Aquilino Ribeiro, n. 18, 1o
3800 Aveiro
Portugal
----------------------------------
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Associated Topics:
College Analysis

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