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MathGroup Archive 2004

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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)
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