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**liquidFuzz** $\displaystyle \displaystyle\sum\limits_{n\geq 1}^\infty \frac{(4iz - 2)^{2n}}{5^n(n+2)}

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The first thing that popped up in my mind was this, absolute value of $\displaystyle \frac{1}{r} = lim |(a_{n+1})/a_n|$. Furthermore, look at start and end points. So I started tinkering with that, but I don't get it...

Convergence radius for n

$\displaystyle \displaystyle \frac { \frac{(z+\frac{i}{2})^{2(n+1)}}{5^{n+1}(n+1+2)}} {\frac{(z+\frac{i}{2})^{2n}}{5^{n}(n+2)}}$ = $\displaystyle \displaystyle \frac { \frac{(z+\frac{i}{2})^{2n} (z+\frac{i}{2})^2}{5^n 5(n+3)}} {\frac{(z+\frac{i}{2})^{2n}}{5^{n}(n+2)}}$ = $\displaystyle \displaystyle \frac { \frac{ (z+\frac{i}{2})^2}{ 5(n+3)}} {\frac{1}{(n+2)}}$ = $\displaystyle \displaystyle (z+\frac{i}{2})^2 \frac{n+2}{5(n+3)}$