Sums and Inequalities
Using the fact that for all integers and real numbers
prove that for all positive integers m and n,
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?
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?
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 is an increasing function of x on the interval [0,1], with integral 1/(n+1).
Originally Posted by hairymclairy
If you divide the unit interval into m equal subintervals, evaluate the function 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.