Let $p_n(z)$ be the $n^\text{th}$ partial sum of the Maclaurin series for $f(z) = (1+z)/(1-z)$. For large $n$ the zeros of $p_n$ appear to avoid the point $z=-1$:

*Figure: Zeros of $p_{40}$ and the unit circle.*

To calculate the asymptotic behavior of zeros like this one can sometimes calculate a scaling limit of the polynomials. For example, consider instead

$$ g(z) = \frac{a+z}{1-z}. $$

The partial sums of $g$, which we'll call $q_n$, are given by

$$ q_n(z) = a + (a+1) z\cdot \frac{1-z^n}{1-z}. $$

Using this expression we can obtain the following formulas:

$$ \lim_{\substack{n \to \infty \\ n\text{ even}}} q_n\!\left(-1+\frac{w}{n}\right) = \frac{a-1}{2} + \frac{a+1}{2} e^{-w}, \tag{1} $$

$$ \lim_{\substack{n \to \infty \\ n\text{ odd}}} q_n\!\left(-1+\frac{w}{n}\right) = \frac{a-1}{2} - \frac{a+1}{2} e^{-w}. \tag{2} $$

So when $a \neq \pm 1$ we can deduce, for example, that the limit points of the zeros of $q_n(-1+w/n)$ for $n$ odd are

$$ w_k := \log\left|\frac{a+1}{a-1}\right| + i\left(\arg \frac{a+1}{a-1} + 2\pi k\right), \qquad k \in \mathbb Z, $$

and hence that the zeros $z_n$ of $q_n(z)$ for $n$ odd which tend to $z=-1$ have the form

$$ z_n = -1 + \frac{w_k}{n} + o(n^{-1}), \qquad k \in \mathbb Z. $$

However, if $a=1$ then these scaling limits $(1)$ and $(2)$ have no zeros, which tells us that the scale $1/n$ is incorrect, and that we should choose another scaling which is $\gg 1/n$. But, carrying out the calculations, it doesn't seem possible to get a limit function with zeros at all.

Is there another approach we could use to find the asymptotics of these zeros near $z=-1$?