I'm guessing the function is
[tex]f(x)=\dfrac5{2-x}=5(2-x)^{-1}[/tex]
whose first few derivatives are
[tex]\dfrac{\mathrm df}{\mathrm dx}=-5(2-x)^{-2}(-1)=5(2-x)^{-2}[/tex]
[tex]\dfrac{\mathrm d^2f}{\mathrm dx^2}=-2\cdot5(2-x)^{-3}(-1)=2!\cdot5(2-x)^{-3}[/tex]
with a general pattern of
[tex]\dfrac{\mathrm d^nf}{\mathrm dx^n}=n!\cdot5(2-x)^{-(n+1)}=\dfrac{5n!}{(2-x)^{n+1}}[/tex]
Then [tex]f[/tex] has power series (centered at 0)
[tex]\displaystyle\sum_{n=0}^\infty\frac{\frac{5n!}{2^{n+1}}}{n!}x^n=\frac52\sum_{n=0}^\infty\left(\frac x2\right)^n[/tex]
which is a geometric series, converging only for
[tex]\left|\dfrac x2\right|<1\implies|x|<2\implies-2<x<2[/tex]
(this is the interval of convergence)
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If you know that for [tex]|x|<1[/tex]
[tex]\dfrac1{1-x}=\displaystyle\sum_{n=0}^\infty x^n[/tex]
then we can derive the same series by manipulating this result:
[tex]\displaystyle\frac5{2-x}=\frac52\frac1{1-\frac x2}=\frac52\sum_{n=0}^\infty\left(\frac x2\right)^n[/tex]
