# Find the power series

z is a complex number.

$$f(z) = \frac{4 + 3z}{(z + 1)(z + 2)^2}$$

$$\frac{1}{1 + z} = \frac{1}{1 - (-z)} = 1 + (-z) + (-z)^2 + \cdots = \sum_{n = 0}^{\infty}(-z)^n$$

$$\frac{1}{(z + 2)^2} = -\frac{d}{dz} \ \frac{1}{1 - (-z - 1)} = -\frac{d}{dz}\sum_{n = 0}^{\infty}(-z - 1)^n = \sum_{n = 0}^{\infty}-n(-z-1)^{n-1}$$

$$f(z) = \sum_{n = 0}^{\infty}-(4 + 3z) n (-z)^n (-z - 1)^{n - 1}$$

Radius of convergence by the ratio test:

$$\lim_{n\to\infty}\left|\frac{-(4+3z)(n+1)(-z)^{n+1}(-z-1)^n}{-(4+3z)n(-z)^n(-z-1)^{n-1}}\right|=\lim_{n\to\infty}\left|\frac{(n+1)(z)(z+1)}{n}\right|$$

$$|z(z+1)|<1$$

Is the above correct?

If so, how do I find the Radius of Convergence from that?

Related Calculus and Beyond Homework Help News on Phys.org
Char. Limit
Gold Member
No, that doesn't look right. For one thing, you can't just combine summands like that. The product of (z+1)(z+2)^2 is going to be the product of the two sums, and you're probably going to end up with a double sum or something.

No, that doesn't look right. For one thing, you can't just combine summands like that. The product of (z+1)(z+2)^2 is going to be the product of the two sums, and you're probably going to end up with a double sum or something.
Then how can I find the power series of that expression?

Char. Limit
Gold Member
I'd split the function up using Partial Fractions. Then you'll get three functions that are easy to devise a power series for.

I'd split the function up using Partial Fractions. Then you'll get three functions that are easy to devise a power series for.
Could I instead use the rule $R\geq\min{R,S} = 1$?

Isn't there a way to do this without the use of partial fractions?

By partial fractions, we can re-write $f(z)$ as
$$\frac{A}{z + 1} + \frac{B}{z + 2} + \frac{C}{(z + 2)^2} = \frac{4 + 3z}{(z + 1)(z + 2)^2}.$$
By direct computations, $A(z + 2)^2 + B(z + 2)(z + 1) + C(z + 1)$ $= Az^2 + 4Az + 4A + Bz^2 + 3Bz + 2B + Cz + C.$
To find the coefficients, we solve the system of linear equations.
$$\begin{pmatrix}1 & 1 & 0 & 0\\ 4 & 3 & 1 & 3\\ 4 & 2 & 1 & 4\end{pmatrix}\Rightarrow \begin{pmatrix}1 & 0 & 0 & 1\\ 0 & 1 & 0 & -1\\ 0 & 0 & 1 & 2\end{pmatrix}$$
So the coefficients are $A = 1$, $B = -1$, and $C = 2$.
Therefore, $\displaystyle\frac{4 + 3z}{(z + 1)(z + 2)^2} = \frac{1}{z + 1} - \frac{1}{z + 2} + \frac{2}{(z + 2)^2}$ $\displaystyle\Rightarrow\frac{1}{1 - (-z)} - \frac{1}{2}\cdot\frac{1}{1 - \left(-\frac{z}{2}\right)} - \frac{d}{dz}\frac{1}{1 - \left(-\frac{z}{2}\right)}$.
The rational expressions can be represented as the following power series
$$\sum_{n = 0}^{\infty}(-z)^n - \frac{1}{2}\sum_{n = 0}^{\infty}\left(-\frac{z}{2}\right)^n - \sum_{n = 0}^{\infty}n\left(-\frac{z}{2}\right)^{n - 1}.$$

Now how do I find the radius of convergence

Char. Limit
Gold Member
First express it as a single sum. Due to associativeness, you can easily rewrite it like this:

$$\sum_{n=0}^{\infty} (-1)^n z^n - (-1)^n z^n 2^{-n-1} - (-1)^n n z^n 2^{1-n}$$

Which, in a simpler form, is:

$$\sum_{n=0}^{\infty} \left( (-1)^n \left(1 - \frac{1}{2^{n+1}} - \frac{n}{2^{n-1}} \right) \right) z^n$$

Now how would you find the radius of convergence of a single sum?

First express it as a single sum. Due to associativeness, you can easily rewrite it like this:

$$\sum_{n=0}^{\infty} (-1)^n z^n - (-1)^n z^n 2^{-n-1} - (-1)^n n z^n 2^{1-n}$$

Which, in a simpler form, is:

$$\sum_{n=0}^{\infty} \left( (-1)^n \left(1 - \frac{1}{2^{n+1}} - \frac{n}{2^{n-1}} \right) \right) z^n$$

Now how would you find the radius of convergence of a single sum?
Ratio test

So we end up getting

$$|z| < 1$$

and R = 1

Correct?

Last edited: