Question

Find the Radius of Convergence and the Interval of Convergence for the series.
SUM x^n/(n3^n) as n=1 to inf

Answer 1

\[\sum_{1}^{\inf} x ^{n}/(n3^{^{n}})\]

Answer 2

ratio test gives me x/3n+3

Answer 3

now from there is where i get lost...as I figure that limit is 0 so I can't figure the radius of convergence or interval.

Answer 4

then you win!

Answer 5

how would i express that though like what would i say ? this is the first time i had a problem like this.

Answer 6

if limit is zero for all \(x\) then radius of convergence is \((-\infty, \infty)\) i.e. it converges for all \(x\)

Answer 7

if the limit was say 3 then the radius of convergence would be \(\frac{1}{3}\)

Answer 8

would the interval of convergence be the same...as I imagine you can't test -inf and inf. so R = AllReal and Interval = ALlLREAL ?

Answer 9

all real yes

Answer 10

but i am not sure your ratio is right. i mean your answer is right, but your ratio may be off

Answer 11

I'll double check it... maybe I messed up some algebra.

Answer 12

\[x ^{n+1}/((n+1)3^{n+1}) * n3^{n}/x ^{n}\]

Answer 13

who scratch that!! the radius of convergence is not \((-\infty, \infty)\) !!!

Answer 14

by was i ever wrong, sorry

Answer 15

oh crap lol...if your wrong then im in big trouble lol.

Answer 16

\[\frac{n3^nx^{n+1}}{(n+1)3^nx^n}=\frac{nx}{3(n+1)}\]

Answer 17

\[\lim_{n\to \infty}\frac{n}{3(n+1)}=\frac{1}{3}\] radius of convergence is 3

Answer 18

Doh! it was my fault, assuming I had the ratio right. I can take it from there. Now I also know what to do in the event if I do have something that = 0.

Answer 19

Thanks satellite73. Your always helping me out and explaining lol :-). I may very well pass this calc course because of you.