Power Series Expansion: Solving for 1/1+z at z=-5 with Radius of Convergence

[Stephen88]

I'm trying to find the power series expansion of 1/1+z at z=-5 and the radius of convergence.How should I think and solve this problem?
I'm looking for a step by step explanation because I want to understand the mechanics behind it.Thank you.

 

[Krizalid1]

This could work: $\dfrac{1}{{1 + z}} = \dfrac{1}{{6 + (z - 5)}} = \dfrac16\cdot\dfrac{1}{{1 + \frac{{z - 5}}{6}}}$

 

[Stephen88]

Thanks but how did you come up with that...and why in this form?

 

[Moo1]

This could work: $\dfrac{1}{{1 + z}} = \dfrac{1}{{6 + (z - 5)}} = \dfrac{1}{{1 + \frac{{z - 5}}{6}}}$

With a $\frac 16$ factor :p

 

[Krizalid1]

Haha, yes, Moo, I forgot to add it.

Stefan because given a power series $\displaystyle\sum_{k=0}^\infty b_k(z-a)^k,$ you have $a$ is the center, so what I'm doing there is to invoke the geometric series centered at $z=5.$

 

[Stephen88]

again same question also is z=-5 meaning it should be z+5 (below) if it was 5 =5 the it should be ...z-5(below),right?

 

[Krizalid1]

I think I get what you mean, so yes.

 

[Stephen88]

so below should be -4+(z+5) maybe?

 

[Krizalid1]

Oh I'm sorry, I misread the question, yes, that's it.
I thought you were asking it for $z=5.$

 

[Stephen88]

so the power series is ...Sum 1/(4^n+1)*(z+5)...with R=-5...?

 

[Prove It]

so the power series is ...Sum 1/(4^n+1)*(z+5)...with R=-5...?

Just to summarise what has been written...

\[ \displaystyle \begin{align*} \frac{1}{1 + z} &= \frac{1}{-4 + (z + 5)} \\ &= -\frac{1}{4}\left[\frac{1}{1 - \frac{z + 5}{4}}\right] \end{align*} \]

and keeping in mind that $\displaystyle \sum_{n = 0}^{\infty}a\,r^n = \frac{a}{1 - r} $ for $ \displaystyle |r| < 1 $, and taking note that $\displaystyle \frac{1}{1 - \frac{z + 5}{4}} $ is of the form $ \displaystyle \frac{a}{1 - r} $, that means

\[ \displaystyle \begin{align*} -\frac{1}{4}\left[\frac{1}{1 - \frac{z + 5}{4}}\right] &= -\frac{1}{4} \sum_{n = 0}^{\infty} \left(\frac{z + 5}{4}\right)^n \end{align*} \]

and this series is convergent where

\[ \displaystyle \begin{align*} \left|\frac{z + 5}{4}\right| &< 1 \\ \frac{|z + 5|}{|4|} &< 1 \\ \frac{|z + 5|}{4} &< 1 \\ |z + 5| &< 4 \end{align*} \]

in other words, the series is convergent inside but not on the circle of radius 4 units centred at z = -5.

 

[Opalg]

A useful check in problems like this is that the radius of convergence is always the distance from the centre of expansion to the nearest singularity. In this case the expansion is centred at $z=-5$, and the function $\frac1{1+z}$ has a singularity at $z=-1$. The distance from –5 to –1 is 4, so that is the radius of convergence (confirming that the answer in the previous comment is correct).