Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2004
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2004

[Date Index] [Thread Index] [Author Index]

Search the Archive

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/


  • References:
    • limits
      • From: homshi@walla.com (homshi)
  • Prev by Date: Re: Re: Piecewise symbol in 5.1
  • Next by Date: Re: Newly Released Mathematica 5.1 Delivers Unmatched Performance for Handling Data
  • Previous by thread: limits
  • Next by thread: Re: Re: limits