Imaginary Common Ratio?

Algebra Level 1

True or false :

$1 + i + i^2 + i^3 + i^4 + \cdots$

The expression above represents an infinite geometric progression sum with first term, $a = 1$ and common ratio $r = i$ . And it can be expressed as $\dfrac a{1-r} =\dfrac 1{1-i} = \dfrac 1{1-i} \cdot \dfrac{1+i}{1+i} = \dfrac{1+i}{1-i^2} = \dfrac12 + \dfrac12 i \; .$

Clarification : $i=\sqrt{-1}$ .

True False

2 solutions

Alan Enrique Ontiveros Salazar
Mar 28, 2016

For the expression to converge we must have $|r|<1$ , but in this case $|r|=1$ , so it doesn't converge.

On the boundary of the taylor series , convergence / divergence has to be determined pointwise.

What we know is:
- If $|r| < 1$ , then the sequence converges
- If $|r| > 1$ , then the sequence diverges
- If $|r| = 1$ , it has to be studied to

As am example, $\sum \frac{ x^n}{n}$ has a convergence radius of 1. On the boundary, it diverges at exactly one point, namely $x = 1$ . It converges on every other point on it's boundary (try to prove that!). With $x = - 1$ , the sum converges to $\ln 2$ .

Calvin Lin Staff - 5 years, 2 months ago

Kay Xspre
Mar 28, 2016

Let the series be $f(x)$ . For $n \in \mathbb{N}$ , the series will be $f(x) = \begin{cases} 1+i & \text{ for } x= 4n+2 \\ i & \text{ for } x= 4n+3 \\ 0 & \text{ for } x= 4n \\ 1 & \text{ fox } x= 4n+1 \end{cases}$ Which means $\displaystyle\lim_{x\to\infty}f(x)$ leads to 4 different values, making the series diverge and not having infinite sum.

Imaginary Common Ratio?

2 solutions

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