Question

Determine if the following series converges/diverges

Answer 1

Here's the quetion itsef. It's an alternating series, I just need help on the second condition of the series, proving that the series (b_n), which they now turn into function of x, is decreasing, namely a_(n+1)<a_n after a certain point.

The problems before it had clear explanations, I feel like they judge fudged this one and at x=0, this thing is not even defined, not the original series, not the b_n and no the derivate of it...

Answer 2

@rational @wio Any help please?

Answer 3

\[
b_{n+1} = \frac{1-(n+1)}{3(n+1)-(n+1)^2} = \frac{1-n-1}{3n+3-n^2+2n+1}
\\ = \frac{(1-n)-1}{(3n-n^2)+2n+4}
\]

Answer 4

Wouldn't the -(n+1)^2 simplify to -(n^2+2n+1)=-n^2-2n-1?

Answer 5

Hmmm, let's see

Answer 6

\[
b_{n+1} = \frac{1-(n+1)}{3(n+1)-(n+1)^2} = \frac{1-n-1}{3n+3-n^2-2n-1}
\\ = \frac{(1-n)-1}{(3n-n^2)-2n+2}
\]

Answer 7

Ok, now what do we do next? And why are some things grouped

Answer 8

We are trying to establish a relationship between the nth and n+1 term

Answer 9

Ok

Answer 10

Okay, so subtract \(b_{n+1}\) from \(b_n\) and see what you result is.

Answer 11

Oh, I'm allowed to do that, ok. I just thought it would always give a value of of just 1 since it's n+1, but that makes sense.

Answer 12

\[\Large \frac{ -n^2+n }{ n^4-6n^3 +3n^2+6n}\]

Answer 13

\[\Large \frac{ -n+1 }{ n^3-6n^2+3n+6}\]

Answer 14

@wio