Taylor´s Inequality ln(1.25) (Calculus II)?

Determine the degree of the Maclaurin polynomial required for the error in the approximation of the function at the value of x to be less than 0.001.

Ln(1.25).

I have that the Absolute value of R sub n (1.25)= ( 1/z^(n+1) )*( (1.25)^(n+1) / (n+1) )

which should be <0.001. But because 0<z<1.25 I don`t get what value should I pick and I don`t know if the value x=1.25 really maximizes

abs(x-0)^(n+1)=abs(1.25-0)^(n+1)= (1.25)^(n+1).

If I were to pick some number z that would give me the maximum on [0,1.25], the first thing that comes to my mind is that I could pick z=0.1,.0001 and keep decreasing the values and the term 1/z will get bigger and bigger. Any ideas? Thanks in Advance and Thank You very Much.!

I think that 1/z^(n+1) can be written as (1/z)^(n+1) am I wrong?

abs()= absolute value.

Re: Taylor´s Inequality ln(1.25) (Calculus II)?

$\displaystyle \ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + ...$

for an alternating series, error magnitude < |first omitted term|