Ratio test (convergent or divergent?)

  • Thread starter freshcoast
  • Start date
  • #1
185
1

Homework Statement




Ʃ n / 2^n
n=1

Homework Equations


ratio test
lim |a(n+1) / a(n)|
n->∞

The Attempt at a Solution



I have the answer and the steps its just there's one part I am confused on,
first I just apply n+1 to all my n terms, which gives me,


Ʃ [(n+1)/2^(n+1)] / [n/2^n]
n=1

and if I multiply the a(n+1) part by the reciprocal of an, I don't understand how the terms 2^(n+1) and 2^n cancel each other out and leaves me with just a 2?
 

Answers and Replies

  • #2
1,985
107
consider rules of exponents (which boils down to multiplication). consider $$x*x=x^2$$ this is the same as $$x^1*x^1=x^{1+1}=x^2$$ thus it seems plausible that $$x^{n+1}=x^n*x^1$$ does this answer your question?
 
  • #3
35,019
6,770

Homework Statement




Ʃ n / 2^n
n=1

Homework Equations


ratio test
lim |a(n+1) / a(n)|
n->∞

The Attempt at a Solution



I have the answer and the steps its just there's one part I am confused on,
first I just apply n+1 to all my n terms, which gives me,


Ʃ [(n+1)/2^(n+1)] / [n/2^n]
n=1
No it doesn't. In the Ratio Test you look at the ratio an+1/an, NOT the sum of that ratio.
and if I multiply the a(n+1) part by the reciprocal of an, I don't understand how the terms 2^(n+1) and 2^n cancel each other out and leaves me with just a 2?
 

Related Threads on Ratio test (convergent or divergent?)

Replies
2
Views
1K
Replies
1
Views
2K
Replies
5
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
17
Views
938
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
6
Views
2K
Top