# Complex geometric series $\frac{1}{6i}\sum_{k=0}^{\infty}\left(\frac{z-3i}{6i}\right)^n$

$$\dfrac{1}{z+3i}$$ 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 $$|z-3i|<5$$

• It looks to me like the sum should be ${1\over 3i-z}$ Is that what you are asking? (I'm assuming that $n$ and $k$ are supposed to be the same letter in the expression for the sum.) Also, why do you say, $|z-3i|<5, z\in\mathbb{C}?$ – saulspatz Nov 23 '18 at 16:40
• Because I need that the centre of the geometric series is $3i$ – pter26 Nov 23 '18 at 17:34
• For the series to converge $|z-3i|<6$. – Yadati Kiran Nov 23 '18 at 17:36
• But a priori |z-3i|<5 in my problem – pter26 Nov 23 '18 at 17:38

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.