Give an approximation for $f(-1)$ with an error margin of less than $0.01$ $f$ is defined by the power series: $f(x) = \sum_{n=1}^{\infty}\frac{x^n}{3^n (n+2)}$ 
I need to find an approximation for $f(-1)$ such that the error margin will be less than $0.01$.
I know I need to use the Taylor remainder and the Laggrange theorem, but I'm not exactly sure how. All the other times I had a function (not a series) and I knew how to calculate. Now I have a series and I don't really understand what to do

$$f(-1)=\sum_{n\geq 1}\frac{(-1)^n}{3^n (n+2)}=\sum_{n\geq 1}\frac{(-1)^n}{3^n}\int_{0}^{1}x^{n+1}\,dx = -\int_{0}^{1}\frac{x^2}{3+x}\,dx \tag{1}$$
hence $f(-1)=\frac{5}{2}-9\log\left(\frac{4}{3}\right)$ and it is enough to find a good approximation of $\log\left(\frac{4}{3}\right)$. Since:
$$ \log\left(\frac{4}{3}\right)=\log\left(\frac{1+\frac{1}{7}}{1-\frac{1}{7}}\right)=2\sum_{n\geq 0}\frac{1}{(2n+1)7^{2n+1}}\tag{2}$$
and the last series converges really fast, we may approximate $\log\left(\frac{4}{3}\right)$ with
$$ 2\sum_{n=0}^{1}\frac{1}{(2n+1)7^n}=\frac{296}{1096} \tag{3}$$
and the approximation error is way less than $\frac{1}{100}$, it is about $3\cdot 10^{-5}$. That leads to:
$$ \boxed{\;f(-1) \approx \color{red}{-\frac{61}{686}}\;} \tag{4}$$
with an approximation error that is about $3\cdot 10^{-3}$.
