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

Just for your curiosity.
As @gammatester answered, you are looking fo $n$ such that $$\frac 1{3^n(n+2)}\lt \frac 1{100}$$ which can rewrite $$3^n(n+2)\gt 100$$ Just by inspection, $n=3$ is the samllest value for which the inequality holds.
In fact, there is an analytical solution to the equation $$x^n(n+k)=\frac 1 \epsilon$$ It is given by $$n=\frac{W\left(\frac{x^k \log (x)}{\epsilon }\right)}{\log (x)}-k$$ where $W(z)$ is Lambert function. As you will see in the Wikipedia page, since, in a case like your, the argument is quite large, you have a good approximation using $$W(z)=L_1-L_2+\frac{L_2}{L_1}+\cdots$$ where $L_1=\log(z)$ and $L_2=\log(L_1)$. Applied to you case $(x=3, k=2,\epsilon=\frac 1 {100})$, this would give, as a real, $$n\approx \frac{5.24544}{\log(3)}-2\approx 2.7746$$
May I suggest you play with this to see how many terms would be required for an error margin of, say, $10^{-6}$ ?
Sooner or later, you will learn that any equation which can write $$A+Bx+C\log(D+Ex)=0$$ has analytical solution in terms of the beautiful Lambert function.
