Ratio test (convergent or divergent?)

freshcoast
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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?
 
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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?
 
freshcoast said:

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.
freshcoast said:
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?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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