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

**nappysnake** · May 26th 2012, 09:33 AM

Hello,

could anybody please help me with the following infinite series .. 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?

**Plato** · May 26th 2012, 09:46 AM

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 $\frac{1}{n^2}~.$

**fqqs** · May 26th 2012, 09:55 AM

Originally Posted by nappysnake

Hello,

could anybody please help me with the following infinite series .. 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?

$\sum_{n=2}^{\infty} \frac{n}{n^{3}-1}$

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

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

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

**nappysnake** · May 27th 2012, 05:36 AM

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

BUT $\frac{n}{n^{3}-1} = \frac{n}{n^{2} ( n-\frac{1}{n^{2}} ) } = \frac{1}{n ( n-\frac{1}{n^{2}} ) } \ge \frac{1}{n^2}$ , $n=2,3,4...$
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?

**Plato** · May 27th 2012, 06:33 AM

Originally Posted by nappysnake

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

$\frac{{\frac{1}{{{n^2}}}}}{{\frac{n}{{{n^3} - 1}}}} = \frac{{{n^3} - 1}}{{{n^3}}} \to 1$

**nappysnake** · May 27th 2012, 06:53 AM

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.

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

LinkBack

Thread Tools

Display

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

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

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

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

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

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

Similar Math Help Forum Discussions

[SOLVED] Fourier series to calculate an infinite series

Sum of an infinite series - comparing with a known series

Does convergence of an infinite series imply convergence of the infinite sequence?

Euler-Maclaurin Series formula on an infinite series?

calculus II areas about axis, pressure, infinite series, MacLaurin series

Search Tags