Geometric Series Convergence and Divergence

[Drakkith]

I'm a little confused on geometric series.

My book says that a geometric series is a series of the type: n=1 to ∞, ∑ar^n-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?

 

[Samy_A]

$\displaystyle \sum_{n=1}^\infty \frac{1}{2^n} = \frac{1}{2} +\frac{1}{4}+\frac{1}{8}+ \cdots =1$
$\displaystyle \sum_{n=1}^\infty \frac{1}{2^{n-1}} =1 +\frac{1}{2} +\frac{1}{4}+\frac{1}{8}+ \cdots=2$

 

[member 587159]

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.

 

[Mark44]

I'm a little confused on geometric series.

My book says that a geometric series is a series of the type: n=1 to ∞, ∑ar^n-1

We have some very nice MathJax notation for this
$$\sum_{n = 1}^{\infty}ar^{n - 1}$$

The script I wrote is \sum_{n = 1}^{\infty}ar^{n - 1}, with a pair of $ characters at the beginning and end.

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?