Taylor expansion of $\dfrac{e^x}{\cos x}$ Let $g\left(x\right) = \dfrac{e^x}{\cos x}$ and find the first four terms of its taylor polynomial. 
The first four terms are: $$1 + x + x^2 + \frac{2x^3}{3} + \frac{x^4}{2}$$ 
Why do you have to solve this problem term by term? Why does just dividing the terms of the series work out? 
Why doesn't the division of the series of $e^x$ and $\cos (x)$ which gives $\sum \limits_{n=0}^{\infty} \dfrac{x^n (2n)!}{n! (-1)^{n}(x^{2n})} $ give the correct series. 
 A: To put it simple: in general
$$\frac{\sum a_n}{\sum b_n}\neq \sum\frac{a_n}{b_n}.$$
A: We probably know that $e^x = 1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \ldots$, and $\cos x = 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} + \ldots$. You say "why does just dividing the terms of the series work out?", and I don't quite know what you mean. But you also ask why we can't divide the nth degree term of the $e^x$ series by the nth degree of the $\cos(x)$ series to get the ending series, and this I can answer.
When we say that $e^x = 1 + x + \ldots$, we really do mean that $e^x$ can be expressed as an infinite sum. And just like finite sums, we cannot divide quotients of sum term-wise. For example, we know that $\dfrac{1 + 1}{2 + 2} = \dfrac{2}{4} = \dfrac{1}{2} \not = \dfrac{1}{2} + \dfrac{1}{2} = 1$. This is why dividing nth terms do not give the correct series.
A: If you want to solve 
$f(x)/g(x) = h(x)$
where $f(x)$ and $g(x)$ are known,
a reasonable way to do it
for general functions
is to write
$f(x) = g(x) h(x)$,
expand the right hand side,
and equate coefficients.
This will give you a recursion
for the coefficients.
Explicitly,
let $f(x) = \sum_{i=0}^{\infty} f_i x^i$
and similarly for $g$ and $h$.
Then
$\begin{align}
\sum_{i=0}^{\infty} f_i x^i
&= f(x) \\
&= g(x)h(x)\\
&=(\sum_{j=0}^{\infty} g_j x^j)
(\sum_{k=0}^{\infty} h_k x^k)\\
&=\sum_{i=0}^{\infty} x^i
\sum_{j=0}^{i} h_j g_{i-j}
\end{align}$
Equating the coefficients of
$x^i$,
$f_i = \sum_{j=0}^{i} h_j g_{i-j}
= h_i g_0 + \sum_{j=0}^{i-i} h_j g_{i-j}
$
so
$h_i = \frac1{g_0}(f_i-\sum_{j=0}^{i-i} h_j g_{i-j})
$.
For more on this type of coefficientology,
look up "generatingfunctionology"
by Herbert Wilf,
freely available for download at
http://www.math.upenn.edu/~wilf/DownldGF.html.
A: I hate denominators. 
First write ${e^x\over\cos(x)}=a_0+a_1x+a_2x^2+a_3x^3+\cdots$ 
Then, since I hate denominators, I consider $e^x=\cos(x)(a_0+a_1x+a_2x^2+a_3x^3+\cdots)$
Let's work from the right hand side, and take an intuitive guess that we only need expand $\cos(x)$ to a third degree polynomial: 
$$\begin{align*}
\cos(x)(a_0+a_1x+a_2x^2+a_3x^3+\cdots) & = (1-x^2/2+\cdots)(a_0+a_1x+a_2x^2+a_3x^3+\cdots) \\
%& = (a_0+a_1x+a_2x^2+a_3x^3+\cdots)-(x^2/2)(a_0+a_1x+a_2x^2+a_3x^3+\cdots)+\cdots\\
& = a_0+a_1x+(a_2-a_0/2)x^2+(a_3-a_2/2)x^3+\cdots\\
& = e^x\\
& = 1+x+x^2/2+x^3/6
\end{align*}$$
Thus, $a_0=1$, $a_1=1$, $a_2=1$, $a_3=2/3$
$${e^x\over\cos(x)}=1+x+x^2+{2\over3}x^3+\cdots$$
So remember, when dividing power series, "I hate denominators". 
