Macluarin series and radius of convergence

[MissP.25_5]

Hello.
I am stuck on this question. I'd appreciate if anyone could help me on how to do this.

The question:
Expand the following into maclaurin series and find its radius of convergence.

$$\frac{2-z}{(1-z)^2}$$

I know that we can use geometric series as geometric series is generally
$$\frac{1}{1-z}=\sum_{n=0}^{\infty}z^n$$

So,
\begin{align*} \frac{2 - z}{ \left( 1 - z \right) ^2 } &= \frac{1 + 1 - z }{ \left( 1 - z \right) ^2 } \\ &= \frac{1}{ \left( 1 - z \right) ^2 } + \frac{1 - z}{ \left( 1 - z \right) ^2 } \\ &= \frac{1}{ \left( 1 - z \right) ^2 } + \frac{1}{1 - z} \end{align*}

$$\frac{d}{dz}\frac{1}{(1-z)}=\frac{1}{(1-z)^2}$$

Now to get the final answer of the maclaurin series expansion, it should start off like this, right?
$$\frac{2-z}{(1-z)^2}= \sum_{n=0}^{\infty}nz^{n-1}+\sum_{n=0}^{\infty}z^n$$

How do I finish this? The answer I was given is :

$$\sum_{n=0}^{\infty}(n+2)z^n$$

 

[HallsofIvy]

You have started off very well. Yes, $\frac{2- z}{(1- z)^2}= \frac{1+ 1- z}{(1- z)^2}= \frac{1}{(1- z)^2}+ \frac{1}{1- z}$. And $\frac{1}{1- z}= \sum z^n$, the "geometric series".

Now, note that $d(1/u)/dx= -1/u^2$. So $1/(1- z)^2= d/dx(1/(1- z))$. Differentiating the geometric sequence "term by term" gives a power series for $1/(1- z)^2$

 

[MissP.25_5]

You have started off very well. Yes, $\frac{2- z}{(1- z)^2}= \frac{1+ 1- z}{(1- z)^2}= \frac{1}{(1- z)^2}+ \frac{1}{1- z}$. And $\frac{1}{1- z}= \sum z^n$, the "geometric series".

Now, note that $d(1/u)/dx= -1/u^2$. So $1/(1- z)^2= d/dx(1/(1- z))$. Differentiating the geometric sequence "term by term" gives a power series for $1/(1- z)^2$

I already did that in my solution. Look,

Now to get the final answer of the maclaurin series expansion, it should start off like this, right?
$$\frac{2-z}{(1-z)^2}= \sum_{n=0}^{\infty}nz^{n-1}+\sum_{n=0}^{\infty}z^n$$

How do I finish this? The answer I was given is :

$$\sum_{n=0}^{\infty}(n+2)z^n$$

 

[HallsofIvy]

I already did that in my solution. Look,

Now to get the final answer of the maclaurin series expansion, it should start off like this, right?
$$\frac{2-z}{(1-z)^2}= \sum_{n=0}^{\infty}nz^{n-1}+\sum_{n=0}^{\infty}z^n$$

You have your first sum starting with $z^{0- 1}= z^{-1}$. Because it is multiplied by n= 0, it doesn't change the sum but I think it is causing confusion. Instead write it as $\sum_{n=1}^\infty nz^{n-1}$ which is, as I said, exactly the same thing.

How do I finish this? The answer I was given is :

$$\sum_{n=0}^{\infty}(n+2)z^n$$

If you meant $$\sum_{n=1}^\infty nz^{n-1}+ \sum_{n=0}^\infty z^n$$
then you need to change the indexing on the first sum. Let j= n- 1 so that n= j+ 1. When n= 1 j= 0. The formula becomes $$\sum_{j= 0}^\infty (j+ 1)z^j+ \sum_{n=0}^\infty z^n$$.
Now use the fact that "j" is just a "dummy index"- replace j with n in the first sum and combine.

 

[MissP.25_5]

You have your first sum starting with $z^{0- 1}= z^{-1}$. Because it is multiplied by n= 0, it doesn't change the sum but I think it is causing confusion. Instead write it as $\sum_{n=1}^\infty nz^{n-1}$ which is, as I said, exactly the same thing.


If you meant $$\sum_{n=1}^\infty nz^{n-1}+ \sum_{n=0}^\infty z^n$$
then you need to change the indexing on the first sum. Let j= n- 1 so that n= j+ 1. When n= 1 j= 0. The formula becomes $$\sum_{j= 0}^\infty (j+ 1)z^j+ \sum_{n=0}^\infty z^n$$.
Now use the fact that "j" is just a "dummy index"- replace j with n in the first sum and combine.

I don't get it, how can I just replace j with n, when j=n-1? But if I do that, then I'll go back to square one.

 

[Ray Vickson]

I don't get it, how can I just replace j with n, when j=n-1? But if I do that, then I'll go back to square one.

You have $S_1 + S_2$, where
$$S_1 = \sum_{n=1}^{\infty} n z^{n-1} \; \text{ and } \; S_2 = \sum_{n=0}^{\infty} z^n$$
In $S_1$ the power of $z$ ranges from $n-1 = 0$ to $\infty$, so if we call that power $j$ (instead of $n-1$) the general term is $(j+1) z^j$, for $j = 0,1,2, \ldots$. Now you can re-name $j$ freely, because you really have something of the form
$$(\text{index-name}+1)\, z^{\text{index-name}}$$
and we are currently using $\text{index-name} = j$ because that is a lot easier to write. However, "index-name" can be anything we want, so we can easily call it $n$ if we want.

If you remain unconvinced, just use the alternative of re-naming the summation index from $n$ to $j$ in the second sum $S_2$ to get
$$\sum_{j=0}^{\infty} (j+1) z^j + \sum_{j=0}^{\infty} z^j .$$
Do you see it now?

 

[MissP.25_5]

You have $S_1 + S_2$, where
$$S_1 = \sum_{n=1}^{\infty} n z^{n-1} \; \text{ and } \; S_2 = \sum_{n=0}^{\infty} z^n$$
In $S_1$ the power of $z$ ranges from $n-1 = 0$ to $\infty$, so if we call that power $j$ (instead of $n-1$) the general term is $(j+1) z^j$, for $j = 0,1,2, \ldots$. Now you can re-name $j$ freely, because you really have something of the form
$$(\text{index-name}+1)\, z^{\text{index-name}}$$
and we are currently using $\text{index-name} = j$ because that is a lot easier to write. However, "index-name" can be anything we want, so we can easily call it $n$ if we want.

If you remain unconvinced, just use the alternative of re-naming the summation index from $n$ to $j$ in the second sum $S_2$ to get
$$\sum_{j=0}^{\infty} (j+1) z^j + \sum_{j=0}^{\infty} z^j .$$
Do you see it now?

Yes, thank you!