# Does this series converge?

• Nov 4th 2008, 04:53 PM
Does this series converge?
Does the series $\displaystyle \sum ^ \infty _{k=1} x^k$ on (0,1)? Pointwise or Uniformly?

Well, so far I have this:

Define the partial sum $\displaystyle S_n = \sum ^n _{k=1} x^k =1+x=x^2+...+x^n= \frac {1-x^{n+1}}{1-x}$

So the series converge point wise to $\displaystyle \frac {1}{1-x}$?
• Nov 4th 2008, 06:28 PM
Mathstud28
Quote:

Originally Posted by tttcomrader
Does the series $\displaystyle \sum ^ \infty _{k=1} x^k$ on (0,1)? Pointwise or Uniformly?

Well, so far I have this:

Define the partial sum $\displaystyle S_n = \sum ^n _{k=1} x^k =1+x=x^2+...+x^n= \frac {1-x^{n+1}}{1-x}$

So the series converge point wise to $\displaystyle \frac {1}{1-x}$?

Just consider $\displaystyle \lim_{x\to{0^-}}\sum_{n=0}^{\infty}x^n$ and use abel's theorem and then you can see that it does not converge unifomrly because it does not equal $\displaystyle \frac{1}{1-0}$
• Nov 5th 2008, 01:10 AM
Opalg
Quote:

Originally Posted by tttcomrader
Does the series $\displaystyle \sum ^ \infty _{k=1} x^k$ on (0,1)? Pointwise or Uniformly?

Well, so far I have this:

Define the partial sum $\displaystyle S_n = \sum ^n _{k=1} x^k =1+x+x^2+...+x^n= \frac {1-x^{n+1}}{1-x}$

So the series converge point wise to $\displaystyle \frac {1}{1-x}$?

The sum starts at k=1, not k=0, so you need to modify the formula for $\displaystyle S_n$ slightly.

To see whether or not the series converges uniformly, you need to think about what happens as $\displaystyle x\to1$.