Sums and Inequalities

• Mar 26th 2011, 09:51 AM
hairymclairy
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(Headbang). Can anyone give me a nudge in the right direction?

Thanks.
• Mar 26th 2011, 12:41 PM
Opalg
Quote:

Originally Posted by hairymclairy
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(Headbang). 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.
• Mar 26th 2011, 12:54 PM
hairymclairy
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?
• Mar 31st 2011, 08:19 AM
hairymclairy
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.
• Mar 31st 2011, 11:48 AM
Opalg
Quote:

Originally Posted by hairymclairy
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.