Recent content by Mathmo

Again thanks for your help, I can show the first part of induction, that for n=1: (1+p) > p which is true. When I try and extend this to n+1 I get: (1+p)^{n+1} > (n+1)p (1+p)^{n}(1+p) > np + p But I don't know how to prove this last step?

-1<r<1 I think I will try to prove this by induction and come back.

Thanks again for your help, but now I'm even more confused: 1) Do you mean p > 0? 2) What allows you to say that (1+p)^n >np

Thanks for your help, but I must just be really stupid, as I'm still stuck on how to progress. :(

\lim_{x\to a}f(x)=L means that given any real number e>0, there exists another real number d, such that: if 0<|x-a|<d then |f(x) - L|<e But I'm not sure what to do next?

The limit is what it tends to, I understand that and know it tends to zero by trying values. e.g. (1/2)^9999, but how do I prove this?

Homework Statement I am trying to prove the sum of a geometric series, but one of the steps involves deriving this result: \lim_{n\to\infty}r^{n}=0 so that you can simplify the sum of a geometric series, where I have got to this stage: S_{\infty} = \frac{a(1-r^{\infty})}{1-r}...