Recent content by euclid3.14

I live in london for all of my life and believe it's the best city in the world :biggrin: The night life is good and there's some really intressting people there and some not so intressting... It's expensive to life there, but the pay you get is most then most other places. You'll love...

I know after you get a phd you get the title of doctor, but how do you become a professor?

Thanks for the help. I've written the sum as (n+2)/ 2(n+1) which tends to 1/2 as n tends to infinity. I also got the odd term to converge to one as p(2n+1) = p2n*a(2n+1) and as p2n tends to 1 and a(2n+1) tends to 1 as well then the odd term tends to one! :bugeye:

Thanks for the help. I've written the sum as (n+2)/ 2(n+1) which tends to 1/2 as n tends to infinity. I also got the odd term to converge to one as p(2n+1) = p2n*a(2n+1) and as p2n tends to 1 and a(2n+1) tends to 1 as well then the odd term tends to one! :bugeye:

Doesn't it tend to one as n tends to infinity as you take the module of the series? There's a chance that I'm confused! could simplify it to \prod_{n=1}^{k}\left((n+1)^2 -1)/(n+1)^2\right or \prod_{n=1}^{k}\left(n^2 +2n /n^2 +2n +1\right) k 1 2 3 4 5 3/4 2/3 5/8 3/5 Seems to be tending to...

Thanks for the responce! I got it to eqaul 1 (think that's right!) I take it that this means the infinite product therefore converges to 1? Have another querry on infinite products. what does \prod_{n=1}^{n}\left(1-1/(n+1)^2\right)? converge to? I've write out the seris to get...

Hi, if an = (-1)^n-1 /n what is the infinite product of (1+an) does working out (1+(a2n-1))(1+a2n) help? :rolleyes:

log(x) = u, u=exp(x), x du = dx. intergral becomes exp[-u] 1/u exp[u] du = 1/u du = log u = log[log(x)] Cheers! :biggrin:

This is doing my head in! I split it to 1/x * 1/log(x) and got the intergral = 1 + the intergral when using intergration by parts. :cry: I know the answer is log[log(x)] but have no idea how you get log of a log. Got a feeling the answer is going to really obvious!