- #1

- 96

- 1

## Homework Statement

Show that the Ʃ 1/(2n-1)^3 Converges

## The Attempt at a Solution

I tried using the ratio the ratio test but that didn't work

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter McAfee
- Start date

- #1

- 96

- 1

Show that the Ʃ 1/(2n-1)^3 Converges

I tried using the ratio the ratio test but that didn't work

- #2

Curious3141

Homework Helper

- 2,850

- 87

I assume the summation is over all non-negative n (i.e. 1,2,3..)?

Just compare with zeta(3) (which has all positive terms and includes all the terms in your series), which converges. If you need to establish convergence of the latter, use the integral test, i.e. prove that ∫x^(-n)dx for the bounds [1, infinity) is finite for all n > 1 (here the n is 3).

EDIT: Damn Latex :grumpy:

Just compare with zeta(3) (which has all positive terms and includes all the terms in your series), which converges. If you need to establish convergence of the latter, use the integral test, i.e. prove that ∫x^(-n)dx for the bounds [1, infinity) is finite for all n > 1 (here the n is 3).

EDIT: Damn Latex :grumpy:

Last edited:

- #3

Dick

Science Advisor

Homework Helper

- 26,263

- 619

I assume the summation is over all non-negative n (i.e. 1,2,3..)?

Just use the comparison test with zeta(3), which converges. If you need to establish convergence of the latter, use the integral test.

Good advice, but it's probably clearer if you say summation 1/n^3 instead of zeta(3). Not everybody knows the Riemann zeta function.

- #4

Curious3141

Homework Helper

- 2,850

- 87

Good advice, but it's probably clearer if you say summation 1/n^3 instead of zeta(3). Not everybody knows the Riemann zeta function.

Fair enough.

- #5

- 96

- 1

Good advice, but it's probably clearer if you say summation 1/n^3 instead of zeta(3). Not everybody knows the Riemann zeta function.

If I say 1/n^3 could I also use the p-series test.

and yes i meant the summation where n=1 to infinity

- #6

Curious3141

Homework Helper

- 2,850

- 87

If I say 1/n^3 could I also use the p-series test.

and yes i meant the summation where n=1 to infinity

If you're allowed to assume the p-series convergence for p = 3, that's essentially the same thing as assuming the convergence for zeta(3). Then you don't even need the integral test. Just compare and you're done.

Share: