I need help understanding the explanation below for how to get limit to a series.?

Question

Note that 1/(3n-1) can be rewritten as 1/(3n - 3 + 2), or
1/(3(n-1) + 2). (This much I can follow of course)

So the series 3∑[ 1/(3n-1) ] from 0 to ∞ is the same as -1 plus the series ∑[ 1/(3n+2) ] from 0 to ∞. (This is the part I don't understand)

Original question: What's the limit of the series(0 to oo) of 9/(3n-1)(3n+2)?

Answer 1

We have that lim (9/(3n-1)(3n+2))/(1/n^2) = lim (9n^2)/((3n -1)(3n+2)). So, we have the limit as n -> oo of the ration2 polynomilas of degree 2. As we know, this limit is just the ratio of the leading coefficientes, so that lim (9/(3n-1)(3n+2))/(1/n^2) = 9/9 = 1. Since 1 >0, the limit comparisson test shows the series ∑9/(3n-1)(3n+2) and ∑1/n^2 are both convergent or both divergent. Since it's well known that ∑1/n^2 converges, it follows ∑9/(3n-1)(3n+2) converges too.

But finding its limit is not easy, I guess we have to resort to numerical methods.

In this case, splitting into 2 rational fractions doesn't help, because you get the difference of 2 series that goes to oo.

Answer 2

you can do a partial fraction to answer that one..split it up into fractions with each factor on the denominator, find you A and B values, and then proceed to get the next step. you can then do the partial sums method to get the sum of "n" terms, and then come up with the sum...its a little hard to do that all on here, seeing i'm not sure of how to use symbols or anything