Finding series of two functions multiplied with each other Let's say we are given a function defined as ${\frac{\ln(1+t)}{1-t}}$. We want to find the series expansion up to ${t^4}$. 
Now we know that we have two function within this larger function which are ${\ln(1+t)}$ and ${\frac{1}{1-t}}$. Now the series and their respective expansions are know and defined as:
$${\ln(1+t)=\sum\limits_{n=0}^{\infty}(-1)^n\frac{x^n}{2}=\frac{1}{2}-\frac{x}{2}+\frac{x^2}{2}-...}$$
$${\frac{1}{1-t}=\sum\limits_{n=0}^{\infty}x^n=1+x+x^2...}$$
Now to find the series expansion as far as ${t^4}$, do we just add or multiply the two series?
 A: You are supposed to multiply the series,
$${\frac{\ln(1+t)}{1-t}}= \frac{1}{1-t}\cdot\ln(1+t)=\sum_{n=0}^{\infty}t^n\cdot \sum_{n=1}^{\infty}(-1)^{n+1}\frac{t^n}{n}.$$
As regards the coefficients of the new series see
the Cauchy product of two power series. 
What do you obtain in this case? What is the series expansion up to $t^4$?
A: Note that:
$$\ln(1+t)=\sum\limits_{n=1}^{\infty}(-1)^{n\color{red}{+1}}\frac{\color{red}t^n}{\color{red}n}=t-\frac{t^2}{2}+\frac{t^3}{3}-\frac{t^4}{4}+O(t^5)\\
\frac{1}{1-t}=\sum\limits_{n=0}^{\infty}\color{red}t^n=1+t+t^2+t^3+O(t^4)\\
$$
Hence:
$$\frac{\ln(1+t)}{1-t}=\frac{1}{1-t}\cdot \ln(1+t)=\\
(1+t+t^2+t^3+O(t^4))\cdot\left(t-\frac{t^2}{2}+\frac{t^3}{3}-\frac{t^4}{4}+O(t^5)\right)=\\
t+\left(t^2-\frac{t^2}{2}\right)+\left(t^3-\frac{t^3}{2}+\frac{t^3}{3}\right)+\left(t^4-\frac{t^4}{2}+\frac{t^4}{3}-\frac{t^4}{4}\right)+O(t^5)=\\
t+\frac{t^2}{2}+\frac{5t^3}{6}+\frac{7t^4}{12}+O(t^5).$$
Alternatively, applying Taylor's formula at $t=0$:
$$\begin{align}y=\frac{\ln(1+t)}{1-t} \Rightarrow (1-t)y=\ln (1+t) &\Rightarrow y(0)=0;\\
-y+(1-t)y'=\frac{1}{1+t} &\Rightarrow y'(0)=1;\\
-2y'+(1-t)y''=-\frac{1}{(1+t)^2} &\Rightarrow y''(0)=1;\\
-3y''+(1-t)y'''=\frac{2}{(1+t)^3} &\Rightarrow y'''(0)=5;\\
-4y'''+(1-t)y^{IV}=-\frac{6}{(1+t)^4} &\Rightarrow y^{IV}(0)=14;\\
\frac{\ln(1+t)}{1-t}=0+\frac{1}{1!}t+\frac{1}{2!}t^2+\frac{5}{3!}t^3+\frac{14}{4!}t^4+O(t^5).\end{align}$$
