Limit of an Infinite SeriesDate: 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. I would very much appreciate your answering this doubt. Yours faithfully, - Juan Jose Cukier - <jjc@infool.ba.ar> Buenos Aires, Argentina. 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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