Srijithju :biggrin:
2*(e - (e-1))
Yikes ! This was the hint g_edgar was tryin to show me, and I didn' know :bugeye:
Oh man, I need to familiarize myself with expansions (...log (1+x))
Thanks Pete K for the hint. Halls and Rob :smile:
Sorry if my original question is misleading... I'm not trying to show that the two series converges, but to what it converges to.
Edit/Added:
And then what I did, instead of getting the number/formula to which it converges, I instead checked for convergence.
Yes, it does converge... but how do you get 2? D Alembert's shows only convergence... what's making me pull my hair off is how to get 2? At first look it seems obvious, I actually ran in Mathematica 7, and yep it is 2. But how?
Another problem
\sum_{r= 1}^\infty \frac{1}{2r(2r+1)}...
THanks srijithju for pointing that out...
it should be:
Σ (r=1→n) (2r-1) / r(r+1)(r+2)
and the formula for getting the sum of n terms for this is n(3n+1)/4(n+1)(n+2)
which I don't know how to get.
Also, my solution in the 1st question is wrong... i was testing for convergence, not...
Thanks edgar.
My initial approach to the 1st question is right, (r+1)!= (r+1) (r) (r-1)... dividing by r, then it all reduces to 1. It's the 2nd one that I can't figure out.
a couple of "Infinite Series" questions...
Hi,
I'm trying to solve some problems in "Stroud's Engg mathematics"...
I'm stuck with these 2 questions:
Σ (r=0→∞) (2r) / (r+1)!
and
Σ (r=0→n) (2r-1) / r(r+1)(r+2)
the 1st question converges to 2. My 1st try is divide everything the...