Re: limits

*To*: mathgroup at smc.vnet.net*Subject*: [mg52333] Re: [mg52309] limits*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Wed, 24 Nov 2004 02:32:04 -0500 (EST)*References*: <200411220628.BAA14179@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

On 22 Nov 2004, at 07:28, homshi wrote: > hello all, > Would you please help me in any of these limits: > Prove: > lim(1^p+2^p+3^p+...+n^p)/n^(p+1)= 1/(p+1) > n->infinity > > lim(1^p+2^p+3^p+...+n^p- n/(1+p))= 0.5 > n->infinity > > lim(1/n)(1+1/2+1/3+...+1/n)=0 > n->infinity > I am not really sure whether I should try to answer this question since: 1. These problems require proofs. This is a Mathematica list and not a math list, and proofs are not something Mathematica does very well at this stage. 2. These look like some kind of homework problems and as a math professor I guess I should disapprove of this method of problem solving. Still the problems were interesting enough to tempt me to solve them so I will give a brief indication how they the first and the last could be solved. The second one, in its present form, seems wrong. 1. I shall assume that p>-1. Consider the integral: Integrate[x^p, {x, 0, 1}, Assumptions -> p > -1] 1/(p + 1) Now compute it using the method of Riemann sums. That is, divide the interval [0,1] into n equal subintervals of length 1/n. The Riemann sum is precisely Sum[(i/n)^p*(1/n), {i, 0, n}] which is just Sum[i^p/n^(p+1),[i,0,n}] Since the limit must be equal to the integral you get your answer. 3. It is enough to show that for every integer k, the sum Sum[1/i, {i,1,n}] can be made less than n/k by taking n large enough. Any such sum looks has the form 1+1/2+1/3 +...+1/k (terms >=1/k) + 1/(k+1) + 1/(k+2)+.... . The first k summands are all greater than 1/k and their sum is less than k. All the remaining terms are less than 1/k with the difference at least 1/(k(k+1)). It it now obvious that if n is large enough, that is if you have enough terms smaller than 1/k, than this total sum can be made less than the sum you would have if all the terms were equal to 1/k, in other words it is less than n/k. Hence n*(n/k) < 1/k. This shows that (1/n)*Sum[1/i, {i,1,n}] is less than 1/k for every positive integer k, provided n is large enough. This proves part 3. 2) Part 2), at least as you stated it is false. We can show it using Mathematica (or equally easily by hand). s[p_] := Sum[1/p^i - n/(1 + p), {i, 1, n}] // Simplify s[p] (((p + 1)*(p^n - 1))/p^n - n^2*(p - 1))/(p^2 - 1) Limit[s[1], n -> Infinity] -Infinity Limit[s[2], n -> Infinity] -Infinity Andrzej Kozlowski Chiba, Japan http://www.akikoz.net/~andrzej/ http://www.mimuw.edu.pl/~akoz/

**Follow-Ups**:**Re: Re: limits***From:*DrBob <drbob@bigfoot.com>

**References**:**limits***From:*homshi@walla.com (homshi)

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