For $\sum_{n=1}^\infty (-1)^n\frac{(x-3)^n}n$ approximate with specific details the series when x = 4, but with an error that does not exceed .01. That is, find a value of n so that the nth partial sum of the series differs from the actual sum of the series by less than .01. (I've already shown that the power series converges for $2 < x \leq 4$ )
I know that $|S - S_n|<a_{n+1}$ So I assume that I need $|S - S_n|<0.01$, therefore, do I just need to find $a_{n+1}$ such that $0.01\leq a_{n+1}$ ?