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[tex]f(z) = \frac{4 + 3z}{(z + 1)(z + 2)^2}[/tex]

[tex]\frac{1}{1 + z} = \frac{1}{1 - (-z)} = 1 + (-z) + (-z)^2 + \cdots = \sum_{n = 0}^{\infty}(-z)^n[/tex]

[tex]\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} [/tex]

[tex]f(z) = \sum_{n = 0}^{\infty}-(4 + 3z) n (-z)^n (-z - 1)^{n - 1}[/tex]

Radius of convergence by the ratio test:

[tex]\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|[/tex]

[tex]|z(z+1)|<1[/tex]

Is the above correct?

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