Re: Re: limits

*To*: mathgroup at smc.vnet.net*Subject*: [mg52391] Re: [mg52333] Re: [mg52309] limits*From*: DrBob <drbob at bigfoot.com>*Date*: Thu, 25 Nov 2004 05:50:34 -0500 (EST)*References*: <200411220628.BAA14179@smc.vnet.net> <200411240732.CAA28746@smc.vnet.net>*Reply-to*: drbob at bigfoot.com*Sender*: owner-wri-mathgroup at wolfram.com

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

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

**Re: limits***From:*Andrzej Kozlowski <akoz@mimuw.edu.pl>

**Re: integration using PSQL algorithm**

**Re: Newly Released Mathematica 5.1 Delivers Unmatched Performance for Handling Data**

**Re: limits**

**Re: limits**