Question

Determine whether the series is convergent or divergent. (1/3)+(1/6)+(1/9)+(1/12)+(1/15)....
I figured out that a(subn) is (1/3)(1/n) but dont know where to go from there...

Answer 1

\[\frac{1}{3}+\frac{1}{6}+\frac{1}{9}+\frac{1}{12}+\frac{1}{15}+\cdots=\sum_{n=1}^\infty \frac{1}{3n}\]
so you're right about the series representation.

To show convergence/divergence, there are a few tests you can use. Do you know the \(p\)-series test? Or the integral test?

Answer 2

i only learned the test for divergence...if it doesnt = 0 then its def divergent but if it does - 0, then i have to take the limit of it

Answer 3

does = 0*

Answer 4

i meant if the limit doesnt = 0, then its def divergent, but if it does, then i have to take the limit and see what i get

Answer 5

my book says that its actually divergent and am trying to figure out why

Answer 6

I was thinking of sequence, sorry about that.

Answer 7

its cool, no worries.

Answer 8

You will use the p-series test on this one.

Answer 9

http://www.youtube.com/watch?v=gB43EA9659c

Answer 10

This guy explains it well

Answer 11

im not familiar with that test. I just learned this stuff today and the only test i learned is the test for divergence where if the limit as n approaches infinity doesnt = 0 then its def divergent, but if it does, then i have to take the limit

Answer 12

ok thanks

Answer 13

You have to find the limit of the sequence of partial sums:
\[\left\{\frac{1}{3},\frac{1}{3}+\frac{1}{6},\frac{1}{3}+\frac{1}{6}+\frac{1}{9},\cdots\right\}_{n=1}^\infty\\
\left\{\frac{12}{36},\frac{18}{36},\frac{22}{36},\frac{25}{36},\cdots\right\}_{n=1}^\infty\]
I can't remember the procedure for finding a formula for the \(n\)th partial sum, unfortunately... But it looks like each successive term increases without bound, and so the series diverges.

Answer 14

ok thanks siths