Sum of Geometric Series by Differentiation

In summary, to find the sum of the series Σ n*(1/2)^n (from n = 1 to n = inf), we first rescaled the sum by factoring out (1/2) and then used the formula Σ r^n (from n = 0 to n = inf) = 1 / (1 - r) if |r| < 1. However, we were stuck when trying to evaluate the first summation. To solve this, we can take the derivative of both sides of Σ r^n (from n = 0 to n = inf) = 1 / (1 - r) with respect to r, which gives us s' = Σ n*r^(
  • #1
mreaume
11
0

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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  • #2
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?
 
  • #3
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.
 

Related to Sum of Geometric Series by Differentiation

What is a geometric series?

A geometric series is a sequence of numbers where each term is multiplied by a constant ratio to get the next term. For example, 1, 2, 4, 8, 16 is a geometric series with a common ratio of 2.

What is the formula for finding the sum of a geometric series?

The formula for finding the sum of a geometric series is S = a / (1-r), where S is the sum, a is the first term in the series, and r is the common ratio between each term.

How do you know if a geometric series is convergent or divergent?

A geometric series is convergent if the absolute value of the common ratio, r, is less than 1. If the absolute value of r is greater than or equal to 1, the series is divergent.

What is the difference between a finite and infinite geometric series?

A finite geometric series has a limited number of terms, while an infinite geometric series continues on infinitely. The sum of a finite geometric series can be calculated using the formula, while the sum of an infinite geometric series can only be approximated.

How is the sum of a geometric series used in real life?

The concept of a geometric series is used in various fields of science and economics, such as compound interest, population growth, and radioactive decay. It can also be used to model natural phenomena, such as the growth of plants and bacteria.

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