Using

\(\displaystyle \frac{1}{1+z} = \frac{1}{3+(z-2)}\)

expand \(\displaystyle \frac{1}{1+z}\) is a Taylor series about z=2.

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Do I just find all the derivatives or something? Perhaps I need to use a geometric series here? Partial fractions? How do I start?

Assume you can write \(\displaystyle \frac{1}{1 + z}\) as a Taylor polynomial.

So \(\displaystyle \frac{1}{3 + (z - 2)} = c_0 + c_1(z - 2) + c_2(z - 2)^2 + c_3(z - 2)^3 + \dots\).

By letting \(\displaystyle z = 2\) we can see that \(\displaystyle c_0 = \frac{1}{3}\).

Differentiate both sides

\(\displaystyle -\frac{1}{[3 + (z - 2)]^2} = c_1 + 2c_2(z - 2) + 3c_3(z - 2)^2 + 4c_4(z - 2)^3 + \dots\).

By letting \(\displaystyle z = 2\) we can see that \(\displaystyle c_1 = -\frac{1}{3^2}\).

Differentiate both sides

\(\displaystyle \frac{2}{[3 + (z - 2)]^3} = 2c_2 + 3\cdot 2c_3(z - 2) + 4\cdot 3c_4(z - 2)^2 + 5\cdot 4c_5(z - 2)^3 + \dots\).

By letting \(\displaystyle z = 2\) we can see that \(\displaystyle c_2 = \frac{1}{3^3}\).

Differentiate both sides

\(\displaystyle -\frac{3\cdot 2}{[3 + (z - 2)]^4} = 3\cdot 2c_3 + 4\cdot 3 \cdot 2c_4(z - 2) + 5\cdot 4 \cdot 3 c_5(z - 2)^2 + 6\cdot 5 \cdot 4 c_6(z - 2)^3 + \dots\).

By letting \(\displaystyle z = 2\) we can see that \(\displaystyle c_3 = -\frac{1}{3^4}\).

Differentiate both sides

\(\displaystyle \frac{4\cdot 3 \cdot 2}{[3 + (z - 2)]^5} = 4 \cdot 3 \cdot 2c_4 + 5 \cdot 4 \cdot 3 \cdot 2c_5(z - 2) + 6\cdot 5 \cdot 4 \cdot 3 c_6 (z - 2)^2 + 7\cdot 6 \cdot 5 \cdot 4 c_7(z - 2)^3 + \dots\).

By letting \(\displaystyle z = 2\) we can see that \(\displaystyle c_4 = \frac{1}{3^5}\).

I think you can see that the series is taking the form

\(\displaystyle \frac{1}{1 + z} = \frac{1}{3} - \frac{1}{3^2}(z - 2) + \frac{1}{3^3}(z - 2)^2 - \frac{1}{3^4}(z - 2)^3 + \frac{1}{3^5}(z - 2)^4 + \dots - \dots\)

\(\displaystyle \frac{1}{1 + z} = \sum_{k = 0}^{\infty}\frac{(-1)^k}{3^{k + 1}}(z - 2)^k\).

Now you need to check the values for which this series will converge.

Since this is a geometric series with common ratio \(\displaystyle r = -\frac{1}{3}(z - 2)\), it only converges for

\(\displaystyle \left|-\frac{1}{3}(z - 2)\right| < 1\)

\(\displaystyle -1 < -\frac{1}{3}(z - 2) < 1\)

\(\displaystyle 3 > z - 2 > -3\)

\(\displaystyle -3 < z - 2 < 3\)

\(\displaystyle -1 < z < 5\).