Ratio test (convergent or divergent?)

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SUMMARY

The discussion focuses on applying the Ratio Test to determine the convergence of the series ∞Ʃ n / 2^n from n=1. Participants clarify that the correct approach involves evaluating the limit of the ratio a(n+1) / a(n) as n approaches infinity. The confusion arises regarding the cancellation of terms 2^(n+1) and 2^n, which simplifies to a factor of 2. This confirms that the series converges, as the limit of the ratio is less than 1.

PREREQUISITES
  • Understanding of the Ratio Test for series convergence
  • Familiarity with limits in calculus
  • Knowledge of exponent rules
  • Basic series notation and summation
NEXT STEPS
  • Review the Ratio Test in detail, focusing on its application to various series
  • Practice evaluating limits involving ratios of sequences
  • Explore the concept of convergence and divergence in infinite series
  • Study additional convergence tests such as the Root Test and Comparison Test
USEFUL FOR

Students studying calculus, particularly those learning about series convergence, as well as educators seeking to clarify the Ratio Test methodology.

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?
 

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