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With a $\frac 16$ factorThis could work: $\dfrac{1}{{1 + z}} = \dfrac{1}{{6 + (z - 5)}} = \dfrac{1}{{1 + \frac{{z - 5}}{6}}}$
Just to summarise what has been written...so the power series is ......Sum 1/(4^n+1)*(z+5)....with R=-5.....?