and the continuous analogue:Originally Posted by TD!
which is easier to prove as we are more familiar with the rules for
manipulating integrals than we are in handling summations.
I remember that quite some time ago, when I first saw this, I didn't believe it was true. Of course, trying for a few values made me see that it was probably true, so all that was left was proving it - now that's up to you
Prove that for all natural n, the following holds:
For those who don't like summation symbols, the above is just math language to say that:
Please don't reveal your answer yet, or post it 'covered'.