I need to find a desired tolerance for a Maclaurin polynomial.

[MonkeySage]

The original function is lnx, where x = 1.25. My error |Rn(1.25)| < 0.001
So I get:
|Rn(1.25)| = |(1/z)(1.25)n+1/(n+1)!|

0<z<1.25

My first attempt with just (1.25)n+1/(n+1) failed, as there is never a point where that will be < 0.001.

[jayem]

My first attempt with just (1.25)n+1/(n+1) failed, as there is never a point where that will be < 0.001.

I think you need to check the original equation again, to see what's missing!!

[AsteriskAmp]

The original function is lnx, where x = 1.25. My error |Rn(1.25)| < 0.001
So I get:
|Rn(1.25)| = |(1/z)(1.25)n+1/(n+1)!|

0<z<1.25

My first attempt with just (1.25)n+1/(n+1) failed, as there is never a point where that will be < 0.001.The formula you've written down for the residue, and the one you said you attempted do not match. You are correct in that the formula you attempted will never satisfy the condition (derivative is strictly increasing, exponential beats linear, or any other argument suffices), the one you originally wrote should do so for z=1 at n=7, and with n=10 be able to do it in all of the interval (not a tight bound, it's probably n=8 but 10 suffices).