Sum of Geometric Series by Differentiation

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SUMMARY

The discussion centers on finding the sum of the series Σ n*(1/2)^n from n = 1 to infinity. The user attempts to rescale the sum and applies the formula Σ r^n = 1 / (1 - r) for |r| < 1. The solution involves differentiating the geometric series with respect to r to derive a formula for Σ n*(1/2)^n. The key insight is that differentiating the geometric series provides a pathway to evaluate the summation effectively.

PREREQUISITES
  • Understanding of geometric series and convergence criteria
  • Familiarity with differentiation techniques in calculus
  • Knowledge of summation notation and manipulation
  • Basic algebraic skills for handling series and limits
NEXT STEPS
  • Learn how to differentiate power series to derive summation formulas
  • Study the application of the formula for Σ r^n in various contexts
  • Explore the concept of series convergence and divergence
  • Investigate advanced techniques in series manipulation and evaluation
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Students studying calculus, mathematicians interested in series summation techniques, and educators teaching geometric series and differentiation methods.

mreaume
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Homework Statement



Find the sum of the following series: Σ n*(1/2)^n (from n = 1 to n = inf).

Homework Equations



I know that Σ r^n (from n = 0 to n = inf) = 1 / (1 - r) if |r| < 1.

The Attempt at a Solution


[/B]
I began by rescaling the sum, i.e.

Σ (n+1)*(1/2)^(n+1) (from n = 0 to n = inf)
= Σ (n+1)*(1/2)*(1/2)^n (from n = 0 to n = inf)
= (1/2) ∑ (n+1)*(1/2)^n (from n = 0 to n = inf)
= (1/2) ( ∑ n*(1/2)^n (from n = 0 to n = inf) + ∑ (1/2)^n (from n = 0 to n = inf) )
= (1/2) ( ∑ n*(1/2)^n (from n = 0 to n = inf) + 1 / (1 - 1/2) )
= (1/2) ( ∑ n*(1/2)^n (from n = 0 to n = inf) + 2 )

I'm stuck here. I don't know how to evaluate the first summation. Any help would be appreciated!
 
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mreaume said:

Homework Statement



Find the sum of the following series: Σ n*(1/2)^n (from n = 1 to n = inf).

Homework Equations



I know that Σ r^n (from n = 0 to n = inf) = 1 / (1 - r) if |r| < 1.

The Attempt at a Solution


[/B]
I began by rescaling the sum, i.e.

Σ (n+1)*(1/2)^(n+1) (from n = 0 to n = inf)
= Σ (n+1)*(1/2)*(1/2)^n (from n = 0 to n = inf)
= (1/2) ∑ (n+1)*(1/2)^n (from n = 0 to n = inf)
= (1/2) ( ∑ n*(1/2)^n (from n = 0 to n = inf) + ∑ (1/2)^n (from n = 0 to n = inf) )
= (1/2) ( ∑ n*(1/2)^n (from n = 0 to n = inf) + 1 / (1 - 1/2) )
= (1/2) ( ∑ n*(1/2)^n (from n = 0 to n = inf) + 2 )

I'm stuck here. I don't know how to evaluate the first summation. Any help would be appreciated!

Try taking the derivative with respect to r of both sides of Σ r^n (from n = 0 to n = inf) = 1 / (1 - r). Can you relate that result to your problem?
 
mreaume said:

Homework Statement



Find the sum of the following series: Σ n*(1/2)^n (from n = 1 to n = inf).

Homework Equations



I know that Σ r^n (from n = 0 to n = inf) = 1 / (1 - r) if |r| < 1.

The Attempt at a Solution


[/B]
I began by rescaling the sum, i.e.

Σ (n+1)*(1/2)^(n+1) (from n = 0 to n = inf)
= Σ (n+1)*(1/2)*(1/2)^n (from n = 0 to n = inf)
= (1/2) ∑ (n+1)*(1/2)^n (from n = 0 to n = inf)
= (1/2) ( ∑ n*(1/2)^n (from n = 0 to n = inf) + ∑ (1/2)^n (from n = 0 to n = inf) )
= (1/2) ( ∑ n*(1/2)^n (from n = 0 to n = inf) + 1 / (1 - 1/2) )
= (1/2) ( ∑ n*(1/2)^n (from n = 0 to n = inf) + 2 )

I'm stuck here. I don't know how to evaluate the first summation. Any help would be appreciated!
Hint: You know ##s = \sum r^n##. Think about differentiating that with respect to ##r## and working with that.
 

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