Question

∑(1to∞)=(k^2+3k+2)/(x^2+x^k+2) use the ratio test to prove that the series converges for x>1

Answer 1

can we first factor the top and bottom to get a simplefied form?

Answer 2

the top can be factored as (k+1)(k+2) but the bottom i cannot factor out

Answer 3

Is this not a power series? The ratio test says:
\[\lim_{k \rightarrow \infty} \left| \frac{((k+1)+1)((k+1)+2)}{x^2+x^{k+1}+2}*\frac{x^2+x^k+2}{(k+1)(k+2)}\right|<1\]
Now take the limit and it should fall out :P

Answer 4

yeah it is not a power series . my only concern was the term with the x^k it wouldn't cancel...

Answer 5

It definitely is a power series :P You use the ratio test to find the radius of convergence for these.

Answer 6

Let me get out some paper and see if i can work this limit out :P

Answer 7

oh. it was given in our instruction that this was not given in a power series. which is why i am confused right now. so if i use the l'hospital's rule to calculate the limit, will the derivative of x^k be k(x)^k-1? or will it be like x^k ln(x)?

Answer 8

Hmmm...h/o just oneee second.

Answer 9

thank you very much!

Answer 10

From what I can tell the derivative of x^k would be: ln(k)x^k

Answer 11

Since you're differentiating with respect to k. However, from there, the limit is exceedingly difficult x.x

Answer 12

ikr? but i think what i was doing wrong the last two nights working on this problem was i am differentiating the x^k wrong..

Answer 13

i somehow see some light because of this clarification..:P

Answer 14

Well if you have:
\[\huge\frac{d}{dx}(a^{u(x)})=\frac{du}{dx}\ln(a)a^{u(x)}\]
Since you're differentiating with respect to k i think that works out that way. But you end up having to do a product rule and it gets grossss.

Answer 15

thanks for your help!

Answer 16

oh thank you very much for that clarification, i somehow calculated the limit to be 1/x which if x >1 the series converges...