From: Ronald Bruck Subject: Re: Sequence of Functions Date: Thu, 28 Dec 2000 11:18:03 -0800 Newsgroups: sci.math Summary: [missing] In article <3ccfet4i0o6r@forum.mathforum.com>, ces_lob@yahoo.com (Cesar Lobo) wrote: > Let > n > ----- > 1 \ x^k - 1 > f_n(x) = ----- ) --------- for all x in (1,Infinity) > x^n / k > ----- > k=1 > > I have the next problems about f_n(x) wich I want to solve (I have > been trying them for a few days) but I donīt know a clear way to solve > them (Iīm a second course math student) so I post this message to see > if someone could solve them or give me a hint. > > 1) How can be evaluated the puntual limit of this sequence? > 2) How can be proved that this sequence converges uniformly? In what is apparently a private e-mail, and not a net post, ces_lob has asked me to explain my earlier post. This should really be done globally, so here goes. First, let me prove that f_n(x) --> 0 as n --> \infty, for any fixed x > 1. Put Hn = 1 + 1/2 + ... + 1/n. Then H_n 1 H_n f_n(x) = ----- ----- \sum_{k=1}^n (1/k) x^k - ----- x^n H_n x^n Since H_n = log(n) + O(1), we see that the H_n/x^n terms approach zero as n --> \infty. The term 1 ----- \sum_{k=1}^n (1/k) x^k H_n is a weighted average of x, x^2, ..., x^n, with the strongest weights on the earlier terms; we should expect, therefore, that this term is something like o(x^{n/2}) or even smaller. So we would PREDICT that the limit is zero. The actual proof requires a little more work. First note that f_{n+1}(x) = (1/x) f_n(x) + (1-x^{-n-1})/(n+1) <= (1/x) f_n(x) + 1/(n+1). Hence, by induction, 1 1 1 1 1 f_{n+p}(x) \le ----- f_n(x) + --------- --- + --------- --- + ... x^p x^{p-1} n+1 x^{p-2} n+2 1 1 1 + --- ------- + ----- x n+p-1 n+p We apply the Cauchy-Schwarz inequality to this last sum, and obtain 1 f_{n+p}(x) <= ----- f_n(x) x^p + \sqrt{1 + x^-2 +...+ x^{-2p-2}} \sqrt{1/(n+1)^2 + ... + 1/(n+p)^2}, hence \limsup_{m \to \infty} f_m(x) <= (1-1/x^2)^{-1/2} \sqrt{1/(n+1)^2 + 1/(n+2)^2 + ...} which converges to 0 as n --> \infty. Now why isn't the limit uniform? Because, if we take z = 1 + 1/n, then z^k = (1 + 1/n)^k >= 1 + k/n by the binomial theorem (or Bernoulli's inequality), hence f_n(z) >= 1/z^n \sum_{k=1}^n (z^k-1)/k >= 1/z^n \sum_{k=1}^n 1/n = 1/z^n. Since 1/z^n --> 1/e, we see that for any 0 < epsilon < 1/e we can NOT force |f_n(x)| < epsilon for all x, and all n sufficiently large: there is always a value of x, namely x = 1 + 1/n for infinitely many n, which will violate this inequality. --Ron Bruck -- Due to University fiscal constraints, all .sigs must be only one line.