Powers and Power Series'

Question

When doing some problems I came across the function: $$f(x)=\frac{x}{1-2x}$$ I realised that the Maclaurin expansion of this function was: $$f(x)=x+2x^2+4x^3+16x^4...$$ Evaluate at $x=1$ to get $$f(1)=1+2+4+16...$$

I have a few of questions about this:

1) By playing around $$g(x)=\frac{x}{1-nx}$$ gives the series for exponents of $n$, is this conjecture true?

2) Is there a way to prove it, any hints?

3) I am aware that the sum of exponents obviously do not converge but why do I get a non-integer (and negative at that) when I evaluate $\frac{x}{1-2x}$ at $x=1$ given that the power series is the sum of integers?

Can a function only be equal to its corresponding power series if it converges for every value of that function?

Bare in mind that being at secondary school I have not covered analysis of any kind so I am woefully incompetent at such topics.

Congratulations--you've found the Generating Function for the sequence $a_n = n^2$. :) — apnorton, Dec 21 '14 at 17:39
Also, I believe that $g(x) = \frac{x}{1-3x}$ (eye-balling, not rigorously checking). — apnorton, Dec 21 '14 at 17:41
Sorry, I made a typo in my question, I conjectured that $\frac{x}{1-nx}$ is the series for the sum of powers of n but I wanted to know how to prove it, I'll amend my question. — Gridley Quayle, Dec 21 '14 at 17:49
Also thanks for that link to generating functions, I'll look into it. — Gridley Quayle, Dec 21 '14 at 17:52

score 1 · Accepted Answer · answered Dec 21 '14 at 17:54

Generally, given a power series $f(x) = \sum_{k-0}^\infty f_k x^k$, there is a radius of convergence $R$ associated with the terms $f_k$. Then if $|x|<R$, the series converges absolutely, if $|x|>R$ then the series diverges and, without knowing anything more, we can say nothing about convergence for $|x|=R$.

The radius of convergence is given by a formula (see here), in the example you have above, we can compute $${1 \over R} = \lim_{n\to \infty} \left| {f_{k+1} \over f_k } \right| = \lim_{n\to \infty}{{1 \over 2} 2^{n+1} \over {1 \over 2} 2^n} = 2$$ And so $R = {1 \over 2}$. Hence the series will diverge at $x=1$.

For 1) For $|x|<1$ we have $1+x+x^2+... = {1 \over 1-x}$, so $x+nx^2+n^2x^3+... = {1 \over n} (nx+ (nx)^2+ (nx)^3+... )= {1 \over 1-nx}$, and this will be valid for $|x| < {1 \over n}$.

This answers 2).

3) The issue here is that $|x|>R$, so while you have some function $f$ and a power series that equals the function for $|x|<R$, you cannot use the series for $|x|>R$. The function $f$ may be perfectly well defined for $|x|>R$ (as in your example) but it does not match its power series based at zero.

Powers and Power Series'

1 Answers1