Results 1 to 5 of 5

Math Help - Sums and Inequalities

  1. #1
    Junior Member
    Joined
    Feb 2009
    From
    London
    Posts
    39
    Thanks
    1

    Sums and Inequalities

    Using the fact that for all integers n\geqslant1 and real numbers t>1

    (n+1)t^n(t-1)>t^{n+1}-1>(n+1)(t-1),

    prove that for all positive integers m and n,

    \frac{m^{n+1}}{n+1}<1^n+2^n+...+m^n<\left(1+\frac{  1}{m}\right)^{n+1}\frac{m^{n+1}}{n+1}.

    I think it has something to do with summing the first inequalities over t, but I've been going round and round in circles with various summations, and can't seem to turn it into the second inequality. Can anyone give me a nudge in the right direction?

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by hairymclairy View Post
    Using the fact that for all integers n\geqslant1 and real numbers t>1

    (n+1)t^n(t-1)>t^{n+1}-1>(n+1)(t-1),

    prove that for all positive integers m and n,

    \frac{m^{n+1}}{n+1}<1^n+2^n+...+m^n<\left(1+\frac{  1}{m}\right)^{n+1}\frac{m^{n+1}}{n+1}.

    I think it has something to do with summing the first inequalities over t, but I've been going round and round in circles with various summations, and can't seem to turn it into the second inequality. Can anyone give me a nudge in the right direction?
    I don't even begin to see how the t-inequalities can be used to prove the m-n-inequalities. But I can see how to prove the m-n-inequalities by integration, using the fact that \int_0^1x^n\,dx = 1/(n+1).

    In fact, \frac1m\sum_{k=1}^m\bigl(\frac km\bigr)^n is an upper Riemann sum for that integral, so that \frac1{n+1}<\frac1m\sum_{k=1}^m\bigl(\frac km\bigr)^n. When you multiply through by m^{n+1}, that becomes \frac{m^{n+1}}{n+1} < \sum_{k=1}^mk^n.

    The other inequality comes in a similar way from the lower Riemann sum obtained by dividing the unit interval into m+1 equal subintervals: \frac1{m+1}\sum_{k=0}^m\bigl(\frac k{m+1}\bigr)^n <\frac1{n+1}.

    But those t-inequalities don't seem helpful at all.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Feb 2009
    From
    London
    Posts
    39
    Thanks
    1
    Ah I see what you mean. I think maybe I'm supposed to use the first identity to show how each series is bounded, and therefore Riemann integrable?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Feb 2009
    From
    London
    Posts
    39
    Thanks
    1
    Hi Opalg,

    I just realised, what is x in your integral in relation to k and m (and t)? I understand everything else, just not how use of that integral is justified as t is greater than 1. Please could you clarify it for me?

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by hairymclairy View Post
    I just realised, what is x in your integral in relation to k and m (and t)? I understand everything else, just not how use of that integral is justified as t is greater than 1. Please could you clarify it for me?
    As I said before, my solution does not use the t-inequalities at all. It depends purely on the fact that x^n is an increasing function of x on the interval [0,1], with integral 1/(n+1).

    If you divide the unit interval into m equal subintervals, evaluate the function x^n at the right-hand end of each subinterval (which is the max value of the function in that subinterval), multiply by the length of the subinterval, and add these m terms, then you get an upper Riemann sum for the function. This sum must be greater than the integral of the function.

    Similarly, you get a lower Riemann sum for the function by dividing the interval into m+1 equal subintervals and evaluating the function at the left-hand end of each subinterval.

    Just to emphasise it again, t does not come into this approach in any way whatever. I totally fail to see how those t-inequalities could help to provide a solution to this problem.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Sums of cos(nx) and sin(nx)
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: November 15th 2010, 10:27 AM
  2. sums
    Posted in the Algebra Forum
    Replies: 2
    Last Post: September 12th 2009, 11:23 AM
  3. sums .
    Posted in the Algebra Forum
    Replies: 1
    Last Post: February 23rd 2009, 01:57 AM
  4. Sums
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: November 1st 2008, 06:30 AM
  5. sums with lnx
    Posted in the Calculus Forum
    Replies: 15
    Last Post: April 21st 2008, 06:11 PM

Search Tags


/mathhelpforum @mathhelpforum