# Converge or Diverge?

1. Apr 26, 2004

### JonF

Does this sum converge of diverge? C is a constant

$\sum_{i=1}^{\infty} x^{C-i}$

Is there an easy way to tell if something converges or diverges?

2. Apr 26, 2004

### arildno

Convergence of the given sum depends on the absolute value of x.

Note that you can rewrite your sum as x^(C)*Sum(i=1,inf)1/(x^(i))
But this is closely related to the well-known geometric series..

3. Apr 26, 2004

### JonF

Ok if X is a whole number, it diverges right?

4. Apr 27, 2004

### matt grime

It is a geometric progression with initial term x^c and common ratio x^{-1} (arildno has a minus sign missing}

as such it converges for all reall x with |1/x|<1, ie |x|>1

5. Apr 27, 2004

### HallsofIvy

Staff Emeritus
Since C is a constant and the sum is over the index i, you can take xC out of the sum and get
$x^C \sum_{i=1}^{\infty} x^{-i}$

You should now be able to recognize the remaining sum as a geometric series in (1/x) which converges for -1< 1/x < 1- that is, x< -1 or x> 1.
Convergence and divergence has nothing whatsoever to do with whether x is a whole number or not.

6. Apr 27, 2004

### JonF

Ok here is my next question, is it possible that:

$X^C < \sum_{i=1}^{\infty} X^{C-i}$

Last edited: Apr 27, 2004
7. Apr 27, 2004

### matt grime

That would depend on what X, and x are. As the sum is $$\frac{x^{C+1}}{1-1/x}$$ given that for the sum to make sense |x|>1. I'm sure you can do the manipulation.

edited to allow for the sum running from 1 to infinity, not 0 to infinity

Last edited: Apr 27, 2004
8. Apr 27, 2004

### arildno

Hmm.., I thought the sum was x^(C)/(x-1)

9. Apr 27, 2004

### matt grime

it could well be, i get bored keeping track of the details. i now think it is x^{c-1}/(1-1/x), which was the second one i put in there and works out at x^c/(x-1) doesn't it?

10. Apr 27, 2004

### arildno

Sure, I thought it was merely a typo or someting.