One can vertify that $1+\frac{1}{2}z^2+\frac{1}{2}z^3$ has no root in the disk $D(0,1.05)$, so $f(z)=\dfrac{1}{1+\frac{1}{2}z^2+\frac{1}{2}z^3}$ should be analytic in $D(0,1.05)$. Assume it has Taylor series: $$f(z)=\sum_{n=0}^\infty c_nz^n,z\in D(0,1.05)$$
On the other hand, note that when $z\in D(0,1)$ , $\frac{1}{2}z^2+\frac{1}{2}z^3\in D(0,1)$ . Since $\dfrac{1}{1+w}=\sum_{n=0}^\infty (-w)^n,w\in D(0,1)$ , we also obtain $$f(z)=\sum_{n=0}^\infty \left(-\frac{1}{2}z^2-\frac{1}{2}z^3\right)^n,z\in D(0,1)\subset D(0,1.05)$$
In $D(0,1)$ , the two series equal to each other, so they should have the same corresponding coefficients. Therefore they should also have the same convergence radius $R$ . By the first series, $R\geq 1.05$ . However, the second series is not convergent at $z=1$, thus $R\leq 1$ . There is a contradiction!
You have here two different series. We have that series $\sum_{n=0}^\infty z^n$ converges in $z=\frac 12,$ but if we do this $$1+\frac 12+\frac 14+ \dots = 1+(1-\frac 12) + (1-\frac 34)+(1-\frac 78) = 2 - \frac 12 + 2 - (\frac 34 + \frac 78) + 2 +\dots.$$ Then we lose convergentity of the series, because every odd term is equal 2($a_n \not\to 0$). That is what you have here. You have series $\sum_{n=1}^\infty a_n z^n,$ then you write $a_n$ in some weird way and sum it in some weird way and you asking yourself why did i lose convergence of series.