Converge or Diverge?

[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?

[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..

[JonF]

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

[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

[HallsofIvy]

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.

[JonF]

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

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

[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

[arildno]

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

[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?

[arildno]

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