# Complex analysis laurent series radius of converence

I am trying to learn laurent series. i almost get it, but i am trying to find out where the following laurent series converges. $$\displaystyle{ \sum_{n=\infty}^{-\infty} \frac{(z^n)}{(3^n+1)} }$$

So first i am looking at $$\displaystyle{ \sum_{n=0}^{\infty} \frac{(z^n)}{(3^n+1)} }$$

and i have got a bit confused finding the radius of convergence!! So i know how to do this but the solution says the radius of convergence is 3, which would have meant the limit was 1/3... how is this possible?? Apologies if this is a stupid question but i am trying to revise for an exam, and maybe a bit panicked!

Also, they have the radius of converence of

$$\displaystyle{ \sum_{n=1}^{\infty} \frac{(w^n)}{(3^{-n}+1)} }$$

as 1, obviously i dont understand how they have got this either!

• yes is that not what i put? Will edit now! – bernardmathews Jun 10 '13 at 22:13
• fixed now - write it out quickly and didnt notice! – bernardmathews Jun 10 '13 at 22:14

For the power series with elements $a_nz^n$ the inverse of the radius of convegence is found via $1/R =\limsup_{n\ge 1}|a_n|^{1/n} = \limsup_{n\ge 1}|1/(3^n+1)|^{1/n}=\limsup_{n\ge 1} 1/(3^n+1)^{1/n}=1/3$, hence the radius is $3$.
In the same spirit $\limsup_{n\ge 1}|1/(3^{-n}+1)|^{1/n} =\limsup_{n\ge 1}|3^n/(3^{ n}+1)|^{1/n}=1$.
$$\lim_{n\to\infty} \left| \frac{z^n}{3^n+1} \right|^{1/n} = \lim_{n\to\infty} \frac{|z|}{(3^n+1)^{1/n}} = \frac{|z|}{3},$$ so the first series converges for $|z|/3 < 1$, i.e. for $|z| < 3$. Similarly $$\lim_{n\to\infty} \left| \frac{w^n}{3^{-n}+1} \right|^{1/n} = \lim_{n\to\infty} \frac{|w|}{(3^{-n}+1)^{1/n}} = \frac{|w|}{1},$$ which shows that the second series converges for $|w| < 1$, i.e. for $|z| > 1$.