Binomial theorem with more than 2 terms

[dyn]

Hi.
Is the binomial theorem $(1+x)^n = 1+nx+(n(n-1)/2)x^2 + ….$ valid for x replaced by an infinite series such as $x+x^2+x^3+...$ with every x in the formula replaced by the infinite series ?

If so , does the modulus of the infinite series have to be less than one for the series to converge ?

 

[fresh_42]

You need to make sure that all $x^n$ converge absolutely.

 

[dyn]

My question arises from an example I have just come across regarding finding the residue of $1/(z^5cos(z))$ about the point z=0. The cos(z) is written out as a power series and then it seems the binomial theorem is applied to it with x replaced by the power series and n= -1. Is the modulus of that power series less than one ?

 

[fresh_42]

What do you mean? $\cos(z)$ is only one series.

 

[dyn]

The example expands $1/coz(z)$ as a binomial of the form $(1+x)^n$ with $x$ represented by the infinite power series starting with $-z^2/2!$ and $n= -1$

 

[fresh_42]

So the question is whether $\dfrac{1}{\lim_{n\to \infty}\sum_{k=0}^n a_k} = \lim_{n \to \infty} \dfrac{1}{\sum_{k=0}^n a_k}$, so what do you know about $\lim_{n \to \infty} \dfrac{f_n}{g_n}$?

 

[dyn]

You've lost me now

 

[fresh_42]

A power series such as $z^5\cos z$ is a limit, as every infinite series. So we have the quotient $1$ divided by that limit of partial sums. You asked whether this can be calculated by as limit of $1$ divided by those partial sums. Write down what you have, with limits instead of $\infty$. This is only a symbol. If you want to know what you can do with it, you have to use its definition.

 

[dyn]

If I am using the binomial expansion to find the residue ie. the coefficient of the $1/z$ term does it even matter if the series converges ? Whether it converges or not I should get the correct coefficient ?