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.

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.
