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MathGroup Archive 2004

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Re: Re: limits


I was confused by your proof for (3), so here goes my attempt.

It's patterned after yours, but I've made the choice of n far more explicit.

Here are some manual steps:

Inequality[Sum[1/i, {i, n}], Equal,
    Sum[1/i, {i, k}] + Sum[1/i, {i, k + 1, n}], Less,
    k + Sum[1/i, {i, k + 1, n}], Equal,
    k + Sum[1/k, {i, k + 1, n}] + Sum[1/i - 1/k,
      {i, k + 1, n}], Less, k + (n - k)/k +
     Sum[1/(k + 1) - 1/k, {i, k + 1, n}], Equal,
    k + (n - k)/k - (n - k)/(k*(k + 1))];

The last term is

sumBound = Last@% // Simplify

(k^2 + n)/(1 + k)

Check to see that sumBound/n is monotone decreasing in n:

Simplify[D[sumBound/n, n]]
-(k^2/((1 + k)*n^2))

Now solve for n:

Solve[sumBound/n == 1/k, n]
{{n -> k^3}}

Hence, for n > k^3, the OP's Sum[...]/n is less than 1/k.

Bobby

On Wed, 24 Nov 2004 02:32:04 -0500 (EST), Andrzej Kozlowski <akoz at mimuw.edu.pl> 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/
>
>
>
>



-- 
DrBob at bigfoot.com
www.eclecticdreams.net


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      • From: homshi@walla.com (homshi)
    • Re: limits
      • From: Andrzej Kozlowski <akoz@mimuw.edu.pl>
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