# Power series from -1 to -infinity?

• Feb 25th 2010, 07:26 PM
davismj
Power series from -1 to -infinity?
http://i49.tinypic.com/zn3903.jpg
I can't really think of any way to prove this. I think it starts with the fact that I'm not really sure what my desired outcome is. I was thinking there might be some way to get it in the form $\displaystyle \sum_{n=0}^{\infty } a_n(z-z_0)^n$. Also, it is very plain to see that $\displaystyle \frac{1}{L}$ is radius of convergence $\displaystyle \sum_{n=0}^{\infty } a_n(z-z_0)^n$, which, it seems, is somehow related since $\displaystyle |z-z_0| < \frac{1}{L} \implies \frac{1}{|z-z_0|} > L$. But...that can't be sufficient...
• Feb 25th 2010, 07:45 PM
Drexel28
Quote:

Originally Posted by davismj
Also, it is very plain to see that $\displaystyle \frac{1}{L}$ is radius of convergence $\displaystyle \sum_{n=0}^{\infty } a_n(z-z_0)^n$, which, it seems, is somehow related since $\displaystyle |z-z_0| < \frac{1}{L} \implies \frac{1}{|z-z_0|} > L$. But...that can't be sufficient...

Really? Why don't you try running with it for a while.
• Feb 25th 2010, 08:31 PM
davismj
Quote:

Originally Posted by Drexel28
Really? Why don't you try running with it for a while.

http://i50.tinypic.com/ohk85s.jpg

Thanks for the tip. Does that look sufficient?
• Feb 25th 2010, 08:38 PM
Drexel28
Quote:

Originally Posted by davismj
http://i47.tinypic.com/9iznt5.jpg

Thanks for the tip. Does that look sufficient?

Right idea, but be careful of what you called $\displaystyle w$. Take one more look!
• Feb 25th 2010, 08:40 PM
davismj
Quote:

Originally Posted by Drexel28
Right idea, but be careful of what you called $\displaystyle w$. Take one more look!

lol, yea that proof was crap. I rewrote it (but its still up above). Thanks for the tip. I was making it way harder than it had to be.