Show that $$|\ln(1+x) - [ x - (x^2)/2 + (x^3)/3 +...+ (-1)^{n-1}((x^n)/n)] | < x^{n+1}/(n+1) $$
For $x \in [0,1]$ and $n \in \Bbb{N}$
I tried starting with the inequality from this question however was unsuccessful as the lefthand side is not always smaller than $\ln(1+x)$ for every n $\epsilon\ \Bbb{N}$. I know that $$|\ln(1+x) - [ x - (x^2)/2 + (x^3)/3 +...+ (-1)^{n-1}((x^n)/n)] | $$ goes to 0 as n goes to infinity, by Taylor's theorem, but I'm not sure if this is useful (I know Taylor comes into play somewhere but thus far haven't been successful in my attempts to incorporate it). I've played around a bit using the exponential function, but I don't think that's the right approach. Any ideas on how to get rolling in proving this? (Should I be thinking induction proof?)
Thanks for your time and any assistance you can provide.
HINT: $$ \ln(1+x)= x - (x^2)/2 + (x^3)/3 +\dots+(-1)^{n-1}((x^n)/n)+\frac{f^{(n+1)}(c)x^{n+1}}{(n+1)!}. $$ By induction you can prove, that $|f^{(n+1)}|$ is of the form $n!$ multiplied by a function less than 1 for $x>0$.