Geometric Series Confusion

In summary, a geometric series is a series of the type ∑ar^n-1, where r is the common ratio and a is the first term. If r<1, the series converges to a/(1-r), otherwise it diverges. When rewriting a series with An = 1/2^n, it becomes a geometric series with r=1/2. This converges to 1/(1-1/2) = 1/(1/2) = 2. However, factoring out a 1/2 changes the series to ∑1/2(1/2)^n-1, which converges to (1/2)/(1-1/2) = (1/
  • #1

Drakkith

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I'm a little confused on geometric series.

My book says that a geometric series is a series of the type: n=1 to ∞, ∑arn-1
If r<1 the series converges to a/(1-r), otherwise the series diverges.

So let's say we have a series: n=1 to ∞, ∑An, with An = 1/2n
An can be re-written as (1/2)n, which apparently makes it a geometric series with r=1/2. This converges to 1/(1-1/2) = 1/(1/2) = 2.

However, I was under the assumption that I was supposed to factor out a 1/2 to make the series ∑1/2(1/2)n-1.
This would converge to (1/2)/(1-1/2) = (1/2)/(1/2) = 1.

Why are these different? Which one is correct?
 
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  • #2
##\displaystyle \sum_{n=1}^\infty \frac{1}{2^n} = \frac{1}{2} +\frac{1}{4}+\frac{1}{8}+ \cdots =1##
##\displaystyle \sum_{n=1}^\infty \frac{1}{2^{n-1}} =1 +\frac{1}{2} +\frac{1}{4}+\frac{1}{8}+ \cdots=2##
 
  • #3
The formule for the total infinite sum is: u1/(1-r)

So, for the first sum, u1 = 1/2, and r = 1/2, so (1/2)/(1-1/2) = 1 and not 2.

So both series are equal.
 
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  • #4
Drakkith said:
I'm a little confused on geometric series.

My book says that a geometric series is a series of the type: n=1 to ∞, ∑arn-1
We have some very nice MathJax notation for this :oldbiggrin:
$$\sum_{n = 1}^{\infty}ar^{n - 1}$$

The script I wrote is \sum_{n = 1}^{\infty}ar^{n - 1}, with a pair of $ characters at the beginning and end.
Drakkith said:
If r<1 the series converges to a/(1-r), otherwise the series diverges.

So let's say we have a series: n=1 to ∞, ∑An, with An = 1/2n
An can be re-written as (1/2)n, which apparently makes it a geometric series with r=1/2. This converges to 1/(1-1/2) = 1/(1/2) = 2.

However, I was under the assumption that I was supposed to factor out a 1/2 to make the series ∑1/2(1/2)n-1.
This would converge to (1/2)/(1-1/2) = (1/2)/(1/2) = 1.

Why are these different? Which one is correct?
 
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1. What is a geometric series?

A geometric series is a sequence of numbers where each subsequent number is found by multiplying the previous number by a constant ratio. For example, the series 2, 6, 18, 54, ... is a geometric series with a common ratio of 3.

2. What is the formula for the sum of a geometric series?

The formula for the sum of a geometric series is S = a(1 - r^n) / (1 - r), where S is the sum, a is the first term, r is the common ratio, and n is the number of terms.

3. How do you know if a series is geometric?

A series is geometric if each term can be found by multiplying the previous term by a constant ratio. Additionally, the ratio between any two consecutive terms should be the same throughout the series.

4. What is the common ratio in a geometric series?

The common ratio in a geometric series is the number by which each term is multiplied to get the next term. It can be found by dividing any term by the previous term.

5. How do you find the sum of an infinite geometric series?

The sum of an infinite geometric series can be found using the formula S = a / (1 - r), where S is the sum, a is the first term, and r is the common ratio. This formula only works if the absolute value of the common ratio is less than 1.

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