Faster way to find the first four non-zero terms of the Maclaurin series for $\frac{1-x}{1+x}\cosh\sqrt{x}$ I want to find the first 4 non-zero terms for :
$$\frac{1-x}{1+x}\cosh\sqrt{x}$$
Before expanding, I rewrite this as $$(1-x)\left(\frac{1}{1+x}\right)\cosh\sqrt{x}$$
Then I expand to get $$(1-x)\left(1-x+x^2-x^3+\cdots\right)\left(1+\frac x {2!}+\frac {x^2} {4!}+\frac{x^3}{6!}+\cdots\right)$$
Now multiplying these brackets and simplifying takes a while (pretty long for my exam time), so is there a faster method to do this or to multiply these brackets?

It's unnecessary to break up $1-x$ and $\frac{1}{1+x}$ into two separate series.  You can just differentiate $\frac{1-x}{1+x}$ directly.  As J.G. pointed out in his comment, it may be easier to rewrite the expression as $\frac{2}{1+x} - 1$ or $2(x + 1)^{-1} - 1$.  This gives you $\frac{1-x}{1+x} = 1 - 2x + 2x^2 - 2x^3 + 2x^4 - 2x^5 + 2x^6 - 2x^7 + ...$.
Then you just have two series to multiply together.
