# Thread: Help with following infinite series: n/[(n^3)-1]

1. ## Help with following infinite series: n/[(n^3)-1]

Hello,

i tried both the ratio test but it is equal to 1 so it is of no help, and the root test is useless, so what can i do to prove it converges? Does it converge?

2. ## Re: Help with following infinite series: n/[(n^3)-1]

Originally Posted by nappysnake;720594 Σ[n/[(n^3)-1
, n=2 .. infinite? i tried both the ratio test but it is equal to 1 so it is of no help, and the root test is useless, so what can i do to prove it converges? Does it converge?
Use the limit comparison test with $\displaystyle \frac{1}{n^2}~.$

3. ## Re: Help with following infinite series: n/[(n^3)-1]

Originally Posted by nappysnake
Hello,

i tried both the ratio test but it is equal to 1 so it is of no help, and the root test is useless, so what can i do to prove it converges? Does it converge?
$\displaystyle \sum_{n=2}^{\infty} \frac{n}{n^{3}-1}$

$\displaystyle \frac{n}{n^{3}-1} = \frac{1}{n^{2} ( n-\frac{1}{n^{2}} ) } \le \frac{1}{n^{2}}$ , $\displaystyle n=2,3,4...$

$\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{2}}$ is known to be convergent, so

$\displaystyle \sum_{n=22}^{\infty} \frac{n}{n^{3}-1}$ also converges.

4. ## Re: Help with following infinite series: n/[(n^3)-1]

Yeah, i thought of the limit comparisson test as my last resource BUT $\displaystyle \frac{n}{n^{3}-1}$ is NOT equal $\displaystyle \frac{1}{n^{2} ( n-\frac{1}{n^{2}} ) }$ , $\displaystyle n=2,3,4...$

BUT $\displaystyle \frac{n}{n^{3}-1} = \frac{n}{n^{2} ( n-\frac{1}{n^{2}} ) } = \frac{1}{n ( n-\frac{1}{n^{2}} ) } \ge \frac{1}{n^2}$ , $\displaystyle n=2,3,4...$
therefor is useless for proving convergence because we require a fraction (series) for that matter which is greater. Am i right? Do i not see something?

5. ## Re: Help with following infinite series: n/[(n^3)-1]

Originally Posted by nappysnake
Yeah, i thought of the limit comparisson test as my last resource BUT $\displaystyle \frac{n}{n^{3}-1}$ is NOT equal $\displaystyle \frac{1}{n^{2} ( n-\frac{1}{n^{2}} ) }$ , $\displaystyle n=2,3,4...$
$\displaystyle \frac{{\frac{1}{{{n^2}}}}}{{\frac{n}{{{n^3} - 1}}}} = \frac{{{n^3} - 1}}{{{n^3}}} \to 1$

6. ## Re: Help with following infinite series: n/[(n^3)-1]

ok i just googled limit comparisson test and i've realized that the limit comparisson test is different than the direct comparisson test and i didn't know that. thank you very much for the help and clarification.