Question

Divergence or Converge?

Answer 1

\[\sum_{n=0}^{\infty}\frac{ 1 }{ 3^{n} - n }\]
what test should i use and why?

Answer 2

comparison test comes to mind

Answer 3

Ratio test to I think

Answer 4

cant say im all familiar with the why tho ... but if we compare it to a function of known convergence ... and it remains 'inside' of it for all n>N ... then it has to converge as well

Answer 5

not ratio test yet because i'm in the limit/comparison test section. the book used the limit comparison test, but i wasn't exactly sure why

Answer 6

we know 1/3^n converges, can we find some value for n such that

1/3^n > 1/(3^n-n)

Answer 7

okay that makes sense. so just to clarify, when using the limit comparison test, i need to find a larger series and evaluate that series in order to determine the status of the original series?
and i can do this by c = lim an/bn?

Answer 8

yes, but the specifics are far removed from me. your book should provide the better steps :)

since they are fractions, the smaller the bottom the bigger the fraction

3^n < 3^n-n

this doesnt work out for 3^n .... so i bet it diverges

Answer 9

http://www.wolframalpha.com/input/?i=sum%28n%3D0+to+inf%29+1%2F%283^n-n%29

my bad :) wolf say it converges ... so that proves im a bit rusty lol

Answer 10

okay, yeah i know it converges because we're comparing it 1/3^n which is a harmonic series and because r = |1/3| < 1 it converges.
so if you're saying we have to choose a larger function to compare it to. but isn't 1/3^n < 1/3n-n?

Answer 11

DIVERGENT

Answer 12

lets say n=1

1/3 > 1/(3-1) is false

n=2

1/9 > 1/(9-2) is false

the right side is going to be larger than the left side .... but then this isnt the LIMIT comparison that im working out.

Answer 13

oh okay thanks. so you're using the comparison test?

Answer 14

http://tutorial.math.lamar.edu/Classes/CalcII/SeriesCompTest.aspx

Paul's got your back ... his example is exactly your question :)

Answer 15

ok thanks!