Question

Determine whether the series is convergent or divergent:

summation n=1 to infinity 1/(n^(3)-n)

I walked through using a similar example as a guide and found the series to be convergent by use of the integral test. Just wanted to make sure this is correct because I only have one submission on this problem

Answer 1

are you sure the index starts at 1 and not 2 ?

Answer 2

because plugging n=1 gives 1/(1^3-1) = 1/0 as first term

Answer 3

Yes, that's why I'm getting hung up. I've done the one with n=2 before, but not n=1

Answer 4

wolfram says sum doesn't exist

http://www.wolframalpha.com/input/?i=summation+n%3D1+to+infinity+1%2F%28n%5E%283%29-n%29

Answer 5

it would be a different story if the index starts from 2

Answer 6

\[\sum_{i=1}^{\infty}\frac{1}{n^3-n}\implies \sum_{i=1}^\infty\left(\frac{1}{n^3}-\frac{1}{n}\right) \]

Answer 7

if n=2, it would be conv - I just didn't know how to handle the n=1

Answer 8

i would listen to wolfram

Answer 9

You went through these? http://www.math.utah.edu/~tskorc/1220/convergence_tests.pdf

Answer 10

Looks kind of like p-series

Answer 11

We've gotten to limit comparison in class, but this problem in in the area where we're doing integral tests

Answer 12

And you haven't learned p-series? That's probably the easiest convergent series tests out of all of them

Answer 13

You're just comparing the powers.

Answer 14

We got to p_series, because limit comparison is when p-series fails

Answer 15

Try p-series.

Answer 16

Using p-series, it would Conv. When I used partial fractions and the integral test, I got 1/2 ln(1) =0. What I don't get is the n=1 because it starts the series with an undefined value.

Answer 17

technically speaking the sum doesn't exists because that undefined term ruins everything

Answer 18

but i feel it is a typo

Answer 19

As do I

Answer 20

I'm going to email my TA about it and see if she has had any other emails about the same problem. Thanks for the help

Answer 21

good luck!