Simplifying a rational function with infinite series in numerator and denominator We're working with Taylor Series and I have to simplify the rational expression
$$ \frac{x-2\frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^5}{5!} - 2\frac{x^6}{6!} + \cdots}{x- \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots}.$$
I'm not sure how to complete the problem -- the answer is 
$$ 1 - x + (x^2/3) - (x^3/6) + \cdots $$
but I'm not sure how to arrive at the solution. Thank you for the help.
 A: The long division mentioned in the comments is likely the way you are 'supposed' to answer the question. Here's the $\operatorname{csc}$ method:
$$\begin{align}R & = \frac{x - 2\frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^5}{5!} - 2 \frac{x^6}{6!} + \ldots}{x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \ldots} \\
& = \operatorname{csc} x \times \sum_{n=0}^\infty \left[\frac{x^{2n+1}}{(2n+1)!}- (1 + [-1]^n) \frac{x^{2n}}{(2n)!}\right] \\
& = \left(\sum_{k=0}^\infty \frac{(-1)^{k-1} 2 (2^{2k-1} - 1) B_{2k}}{(2k)!} x^{2k-1}\right)\left(\sum_{n=0}^\infty \left[\frac{x^{2n+1}}{(2n+1)!}- (1 + [-1]^n) \frac{x^{2n}}{(2n)!}\right]\right) \\
& = \sum_{n,k=0}^\infty\left[\left(\frac{(-1)^{k-1} 2 (2^{2k-1} - 1) B_{2k}}{(2k)!}\right)\left(\frac{1}{(2n+1)!}\right) x^{2(n+k)} \right.\\
& \hphantom{= \sum_{n,k=0}^\infty} \left.- \left(\frac{(-1)^{k-1} 2 (2^{2k-1} - 1) B_{2k}}{(2k)!}\right) \left(\frac{1-[-1]^n}{(2n)!}\right)x^{2(n+k)-1} \right] .\end{align}$$
Renumbering to group the coefficients of particular powers of $x$ using $k = j - n$:
$$\begin{align}R& = \sum_{j=0}^\infty \left[ \left\{ \sum_{n=0}^j \left(\frac{(-1)^{j-n-1} 2 (2^{2[j-n]-1} - 1) B_{2[j-n]}}{(2[j-n])!}\right)\left(\frac{1}{(2n+1)!}\right)\right\} x^{2j} \right.\\
& \hphantom{= \sum_{n,k=0}^\infty} \left.- \left\{ \sum_{n=0}^j \left(\frac{(-1)^{j-n-1} 2 (2^{2[j-n]-1} - 1) B_{2[j-n]}}{(2[j-n])!}\right) \left(\frac{1-[-1]^n}{(2n)!}\right) \right\} x^{2j-1} \right].\end{align}$$ Granted, the expression is pretty complicated, and could do with some simplification, particularly the sums in the curly braces. Assuming I didn't make any mistakes, though, the expressions are correct.
