Re: limits

*To*: mathgroup at smc.vnet.net*Subject*: [mg52364] Re: [mg52309] limits*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Wed, 24 Nov 2004 02:32:40 -0500 (EST)*References*: <200411220628.BAA14179@smc.vnet.net> <C43EC797-3D32-11D9-AFA9-000A95B4967A@mimuw.edu.pl> <73D1FB1A-3D8F-11D9-AFA9-000A95B4967A@mimuw.edu.pl>*Sender*: owner-wri-mathgroup at wolfram.com

I was wrong again (all this hurry!) but the conclusion is still the same, statement 2 is false. In[1]:= s[p_] := Simplify[Sum[i^p, {i, 1, n}] - n/(1 + p)] In[2]:= s[1] Out[2]= n^2/2 In[3]:= s[2] Out[3]= (1/6)*n*(2*n^2 + 3*n - 1) In[4]:= Limit[s[1], n -> Infinity] Out[4]= Infinity In[5]:= Limit[s[2], n -> Infinity] Out[5]= Infinity Andrzej On 23 Nov 2004, at 21:37, Andrzej Kozlowski wrote: > There was an error in my reply, which however makes no difference to > the final outcome. > The correct definition of s in problem 2 is: > > s[p_] := Simplify[Sum[1/i^p - n/(1 + p), {i, 1, n}]] > > which gives: > > s[p] > > > ((p + 1)*HarmonicNumber[n, p] - n^2)/(p + 1) > > With that: > > Limit[s[1], n -> Infinity] > > -Infinity > > and > > Limit[s[2], n -> Infinity] > > > -Infinity > > Which in any case shows that the statement in part 2 is false. > > On 23 Nov 2004, at 10:33, Andrzej Kozlowski wrote: > >> >> 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/ >> >

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