Recent content by yamborghini

Well considering I've been doing this problem for 2 days I think I'm missing something from my tool box. Could you give me a clue? The furthest I've gotten is to factor the 2^p out and then simply put everything in brackets to the power of n. I'm not even sure if this is the right way to go...

I shall omit the modulus signs and the lim in the equations. 2^{p(n+1)}((n+1)+p)!\over((n+1)+p)^{(n+1)} multiplied by (n+p)^n\over 2^{pn}(n+p)! 2^{pn}2^{p}(n+1+p)(n+p)!(n+p)^n \over (n+1+p)^n(n+1+p)2^{pn}(n+p)! then cancelling gives me 2^{p}(n+p)^n \over (n+1+p)^n Is there...

Hi BvU, I've checked it. Got the sheet in front of me and it is definitely that. (unless you mean I have something wrong in my attempt. )

Homework Statement For which integer values of p does the following series converge: \sum_{n=|p|}^{∞}{2^{pn} (n+p)! \over(n+p)^n} Homework Equations The Attempt at a Solution I'm trying to apply the generalised ratio test but get down to this stage where I'm not sure what...