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/

```

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• From: homshi@walla.com (homshi)
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