Is it true that $\sum_{n=0}^{\infty} a_n \frac{(z_0+h)^n-z_0^n}{h} = \sum_{n=1}^{\infty} a_n \frac{(z_0+h)^n-z_0^n}{h}$ (Notice the index)

Question

There's a proof in my textbook where the author uses the followng equality,

$$ \sum_{n=0}^{\infty} a_n \frac{(z_0+h)^n-z_0^n}{h}- \sum_{n=1}^{\infty}na_n z_0^{n-1} = \sum_{n=1}^{\infty} a_n \left(\frac{(z_0+h)^n-z_0^n}{h} - nz_0^{n-1}\right) $$

where $z_0,h$ are fixed values in some appropriate subset of $\mathbb{C}$. The details don't matter for my question, the point is that the two series are both convergent.

The above equality doesn't seem justified to me

The important thing here is not the specific summands, but the change of index in one of the series: in order for the above equality to hold we seem to need that

$$ \sum_{n=0}^{\infty} a_n \frac{(z_0+h)^n-z_0^n}{h} = \sum_{n=1}^{\infty} a_n \frac{(z_0+h)^n-z_0^n}{h} $$

before we can combine them, however this doesn't really seem to be true?

For example it isn't true that $\sum_{n=0}^{\infty} (\frac{1}{2})^n=\sum_{n=1}^{\infty} (\frac{1}{2})^n$

Am I missing something or is the author of the proof not justified in doing so?

Answer 1

It is understood that $z^{n}=1$ when $n=0$. So $(z+h)^{n}-z^{n}=1-1=0$ so the term corresponding to $n=0$ vanishes.

Answer 2

When $n=0$, the summand in the first series is simply $0$.