Respuesta :

LammettHash
[tex]1+\dfrac z5+\dfrac{z^2}{25}+\dfrac{z^3}{125}+\cdots=\displaystyle\sum_{k=0}^\infty\left(\frac z5\right)^k[/tex]
[tex]=\dfrac1{1-\frac z5}=\dfrac5{5-z}[/tex]

provided that [tex]\left|\dfrac z5\right|<1\iff|z|<5[/tex].