Prove that for any positive integer ,

Recall the weighted mean value theorem:

For functions and continuous on , if never changes sign in then there exists such that

* Proof. * In order to apply the weighted mean value theorem (Theorem 3.16 of Apostol) we need to identify the functions and of the theorem. So, let

Then,

Furthermore, for a positive integer , does not change sign on the interval (since and do not change sign on this interval). So we may apply the theorem to conclude there exists such that

Now we need to evaluate this integral. (Here I’m going to cheat a little. In the book we do not yet have techniques available to evaluate this integral. We’ll develop the Fundamental Theorem of Calculus, etc, in Chapter 5 that will allow us to do this. If you have a way to do this without directly evaluating an integral that we don’t yet know how to evaluate * please * do leave a comment with your solution.)

Where the final equality follows since if is even and equals if is odd. Then multiplying that by the , we get .

Putting this together we then have,

Hi,

Your algorithm is correct but a simpler solution could be by using substitution and then by parts. The answer comes out to be a simple one!

Thanks!

The preface to the 2nd edition says, “…the mean-value theorems and routine applications are introduced at an earlier stage.” I bet in the original edition, this problem was in a section

after the fundamental theorem of calculus was introduced.

After spending upwards 10 hours trying to solve this without invoking a fundamental theorem, I’m glad to have found I’m not missing anything obvious.

Yeah, I spent a long time trying to find a way without any success. There might be a way, but whatever it is, I don’t think it’s obvious.

I think it has to do with the telescoping property for product notation.