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
 A: You can solve this without Taylor series. Observe that
$$f(-1) = \sum_{n=1}^\infty \frac{(-1)^n}{3^n(n+2)}.$$ This is a convergent alternating series with $a_n=\frac{1}{3^n(n+2)}$ and you can use the error term
$$\Big|f(-1) -\sum_{n=1}^m (-1)^n a_n\Big| \le a_{m+1}$$
Since $a_3= 1/135 \approx  0.0074\;$ you have 
the approximation $$f(-1) = -\frac{1}{9}+\frac{1}{36} = -\frac{1}{12}$$
A: Let us find a closed form for $H(x) = \sum_{n=1}^{\infty}\frac{x^n}{(n+2)}$. Note that  $$x^2 H(x) \equiv G(x) = \sum_{n=1}^{\infty}\frac{x^{n+2}}{n+2}$$. 
We proceed to find $G(x)$ in closed from. Then, your answer is $f(-1)=H(-1/3)=G(-1/3)/(-1/3)^2$.
Note that: $G'(x) =  \sum_{n=1}^{\infty}x^{n+1}$ and $G(0)=0$. Hence, (for $\vert x\vert <= 1$) we just sum up a geometric series: $$\frac{dG}{dx}(x)=\frac{x^2}{1-x} \quad \cal{and} \quad G(0)=0.$$
The solution reads $$G(x)=-(x+\frac{x^2}{2}+\ln\vert1-x\vert).$$
To conclude this, the final answer is $9 G(-1/3) = -8.913865206602750 \times 10^{-2}$ (confirmed numerically also).
A: 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.
A: $$f(-1)=\sum_{n\geq 1}\frac{(-1)^n}{3^n (n+2)}=\sum_{n\geq 1}\frac{(-1)^n}{3^n}\int_{0}^{1}x^{n+1}\,dx = -\int_{0}^{1}\frac{x^2}{3+x}\,dx \tag{1}$$
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 $3\cdot 10^{-3}$.
