First test of divergence lim n-> n / 8^n

[Kyousik]

Homework Statement

n / 8^n

Homework Equations

The Attempt at a Solution

It converges to 8 / 49? Not sure how.

First test of divergence
lim n-> n / 8^n. infinity / infinity = 1. BUT bottom grows fast.
Using L`Hospital
lim n -> 1 / 3*8^n*ln(2) ---> goes to 0

Tried to use the ratio test

[ 1 / 3*8^(n+1)*ln(2) ] * [ 3*8^n*ln(2) ]
1/8 is left from the ratio test

 

[lalbatros]

What do you mean by "n / 8^n".
This is not a clear problem statement.
And how did you come to guessing the solution as 8/49?

Do you mean you have to calculate Sum[n / 8^n,{n,0,Infinity}] ?
If this is what you mean, then indeed 8/49 is the correct answer.
(I did more effort to guess what you meant than yourself to explain your need)

My hint:

Sum[1 / 8^n,{n,0,Infinity}] is an easy series
Sum[1 / 8^(x*n),{n,0,Infinity}] is just as easy
Sum[n / 8^(x*n),{n,0,Infinity}] is related to the previous one by a simple operation
Sum[n / 8^n,{n,0,Infinity}] is a special case of the previous

 

[Hurkyl]

My hint:

Sum[1 / 8^n,{n,0,Infinity}] is an easy series
Sum[1 / 8^(x*n),{n,0,Infinity}] is just as easy
Sum[n / 8^(x*n),{n,0,Infinity}] is related to the previous one by a simple operation
Sum[n / 8^n,{n,0,Infinity}] is a special case of the previous

That's cute. I would have used (x/8)^n, myself.

 

[sutupidmath]

this series obviously converges.
Here is a little analysis on how to come to this conclusion.
Are you familiar with Dallambers covergence test( i am not sure if it is spelled dallamber, however)?
It states that if limn->infinity a_(n+1)/a_n is larger than 1 it diverges, if it is smaller than 1, and positive it converges. so

lim n->infinitya_n+1/a_n=lim n->infinity[8^n(n+1)]/[8n*8^n]=lim n->infinity(n+1)/8n=1/8<1, so the series converges

 

[lalbatros]

d'Alembert's ratio test
see wiki