Recent content by ironman

Yes, I do. Thanks for the help all!

Oh well... I blame it on my lack of sleep, ha. So I end up with polynomials only. 4-degree in the denominator and 3-degree in the numerator. (highest degrees)

I've no idea how to do that...

Hmm I see. You would get (n+1) * (2n+1) * (2n+2) * (4n)! / (4n +4)! Can you divide both the nom. and denom. by (4n+4)! and end up with 0/1 = 0 as [ n -> ∞ ] ? Weird that my calculus book leaves the |x +c| out then...

Ah of course! (n+1)!/ n! = n+1. I still don't know these rules. But I used : n=3 4*3*2*1/3*2*1 = 4 = n+1

Ah ok I'm sorry. Yes indeed. But you would get: So you might as well just use the 'n-part', for the (x-c) is just a number (?) I ended up with: It's not possible to simplify this, right?

Homework Statement [/B] I have to find the radius of convergence and convergence interval. So for what x's the series converge. The answer is supposed to be for every real number. So the interval is: (-∞, ∞). So that must mean that the limit L = 0. So the radius of convergence [ which is...

Ah the limit is now 1 . So: Limit < infinity. And ##(2/3)^n## converges. So the original function must converge as well.

Homework Statement Show if this sequence (with n=1 to infinity) diverge or converge Homework Equations [/B] The Attempt at a Solution If I use the Limit Comparison Test: compare with so you get that equals lim n -> inf => inf. Can I use the Test like this? What does this...

I got the answer! the first 1/2 (expansion from e^3x) is supposed to be 4 1/2 not 1/2, because you get (3x)^2 / 2! not (x)^2/2!

Homework Statement [/B] lim x -> 0 2. Homework Equations Taylor series for sin cos e and ln () The Attempt at a Solution I tried expanding the sine to 3-degree, and everything else 2-degree. I ended up with this: Now the problem is that WolframAlpha says it should be -6/25. Now if...

Ok thanks Mark44 :) And thanks for the help guys!

I'm sorry. I mean:

Than it must be 3^x... but (ln(x))^3 with x to infinity equals: e to the power something equals infinity therefore (e^∞)^3 which grows faster than 3^∞ , right?

Ahh now it makes sense I think. So in the nominator 3^x is the fastest growing term, and in de denominator that would be ln(x)^3 (?) So if you divide by those terms, both the fastest growing terms will become 1 and the others will become 0.