$\dfrac{1}{z+3i}$ can be interpreted as the sum of the geometric series $\displaystyle\dfrac{1}{6i}\sum_{k=0}^{\infty}\left(\frac{z-3i}{6i}\right)^n$ this can be obtained by writing: $\dfrac{1}{z+3i}=\dfrac{1}{6i+(z-3i)}=\dfrac{1}{6i}\dfrac{1}{1+\dfrac{1}{6i}(z-3i)}$ now the book I'm reading says that this is equal to the series I said at the beginning. But how is it possible if the sum of a geometric series is in the form of $\dfrac{1}{1-q}$?
Edit: I am sure there is convergence of the geometric sum because $|z-3i|<5$
Probably I don't fully understand your question. But let's make a try.
Consider the following series:
$$S = q + qx + qx^2 + qx^3 + \ldots+qx^{n-1} $$
Now let's multiply the whole series by $x$:
$$Sx = qx + qx^2 + qx^3 + qx^4 + \ldots+qx^n$$
Now let's subtract the second from the first:
$$S - Sx = (q + qx + qx^2 + qx^3 + \ldots+qx^{n-1}) - (qx + qx^2 + qx^3 + qx^4 + \ldots+qx^n)$$
We can easily show that only few terms survive at the right side, that is,
$$S(1 - x) = q - qx^n$$
Whence
$$S(1-x) = q(1-x^n)$$
$$S = q\frac{1-x^n}{1-x}$$
As $n$ goes to infinity, the absolute value of $x$ must be less than $1$ for the series to converge, the sum then becomes
$$q + qx + qx^2 + qx^3 + \ldots = \sum_{k = 0}^{+\infty} qx^k = \frac{q}{1-x} ~~~~~~~ |x| < 1$$
And for $q = 1$ we just get
$$\frac{1}{1-x}$$
As wanted.