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Limit of an integral

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Limit of an integral

In a physical problem, I got the following integral:

I = integral(x=0..1) { (x^n - a^n) / (x - a) } dx with a = 1/2

I know, from physical conditions, that this integral must go to 0 when n goes to infinity, can somebody prove it? Could it be true for all values of a between 0 and 1?

bob1, how you show that
1/n+a/(n-1)+...+a^{n-2}/2+a^{n-1} --> 0
when n --> \infty?

I have considered the way that in the integrand one could utilize the mean value theorem to the function x^n.


First, thanks for trying to help (Bob1, Nathan and Pahio). But I still have a problem.
Curiously, you suggest exactly the reverse way I did it: I first got the sum:

R_n = Sum(m=1..n) { (2^(m-n)) / m }

Then I put it into an integral form, thinking that it should be easier to handle.

R_n = I_n = integral(x=0..1) { (x^n - a^n) / (x - a) } dx (a = 1/2)

You seem to think that the sum is easier to handle, and may be you are right. The problem is that the size of the sum n, is also part of the general term. The fact that this general term approches zero, as noted by Nathan, doesn't mean nothing, it is not even a sufficient convergence condition.

From the sum definition of R_n, it is easy to prove the following recursive relation:

R_(n+1) = R_n/2 + 1/(n+1)

Does this help? I remind that the challenge is to prove that R_n (or I_n) goes to zero when n goes to infinity.

That recursive relation does help.

First, notice that if the sequence has a limit, call it R, then the recursive relation tells you that (having taken limits of both sides):

R=R/2+0, i.e. R=0.

All that is left is to show that your sequence converges. There are many ways to do this, but if you'd like to use your recursive relation some more, I'd point out that R_2=1<2, and that you can use your recursive relation to show that if R_n <6/(n+1), then R_{n+1} <6/(n+2).
I hope this helps.

You split the sum into 2 pieces. First from k=0 to \sqrt{n} . Bound this sum by \sqrt{n}/(n-\sqrt{n}). Then take the sum from \sqrt{n} to n. Bound this sum by \sum_{k=\sqrt{n}}^n a^k which is the tail of a convergent series.

Great! Problem solved.

This isn't so bad if you first perform the long division in the integrand and get a polynomial. Then integrate. After that it's a calculus problem.


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