Coefficients in the power series

[esisk]

Hi,

I am trying to prove something, but I need some kind of a result on the coefficients of a power series.

Suppose f(z) has a power series expansion about zero (converges). What can I say about the sum of the absolute values of the coefficients? Ideally I would like to show this sum is less than 1.

thank you

 

[l'Hôpital]

$$\frac{1}{1 - x} = \sum_{n=0}^{\infty} x^n for \ all \ | x | < 1$$

If we denote a_n to be the coefficient of the nth term, then then a_n = 1 for all n, in which case,

$$\sum_{n = 1}^{\infty} 1 \rightarrow \infty$$

Or in the case when it does converge, consider:

$$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$$

Letting x = 1, we'd be summing the coefficients, i.e.

$$\sum_{n=0}^{\infty} \frac{1}{n!} = e$$

It seems your hypothesis fails.

 

[esisk]

Thank you L'Hospital, that is very clear

 

[esisk]

In light of L'Hospital's comment I need change my plan of of attack on the question. I think I need to state the question I was trying to prove:

If f(z) : D--->D where D is the open unit disk, and

the first (k-1) derivatives at zero vanish,

I would like to show that

If(z)I <= IzI^k Tha is abs{f(z)} \leq abs{z}^k

I believe one can (an the question is possibly intended to be solved this way) approach it using Schwarz, I was trying to do it by brute force whereby expanding the power series and saying
f(x)=z^k g(z) and somehow show that I g(z) I <= 1. I am failing as of now.
Any help? I thank you for your time