You have
$$F(x)=\frac{x}{\sum_{n\ge 1}\frac{(-1)^{n+1}}nx^n}=\frac1{\sum_{n\ge 0}{\frac{(-1)^n}{n+1}}x^n}=\frac1{1-\frac{x}2+\frac{x^2}3-+\dots}\;.$$
Suppose that $F(x)=\sum\limits_{n\ge 0}a_nx^n$; then you want
$$1=\left(1-\frac{x}2+\frac{x^2}3-+\dots\right)\left(a_0+a_1x+a_2x^2+\dots\right)\;.$$
Multiply out and equate coefficients:
$$\begin{align*}
1&=a_0\,;\\
0&=a_1-\frac{a_0}2=a_1-\frac12,\text{ so }a_1=\frac12\,;\\
0&=a_2-\frac{a_1}2+\frac{a_0}3=a_2-\frac14+\frac13=a_2+\frac1{12},\text{ so }a_2=-\frac1{12}\,;
\end{align*}$$
and so on. In general $$a_n=\frac{a_{n-1}}2-\frac{a_{n-2}}3+\dots+(-1)^{n+1}\frac{a_0}{n+1}$$ for $n>0$, so
$$\begin{align*}&a_3=-\frac1{24}-\frac16+\frac14=\frac1{24}\;,\\
&a_4=\frac1{48}+\frac1{36}+\frac18-\frac15=-\frac{19}{720}\;,\\
&a_5=-\frac{19}{1440}-\frac1{72}+\frac1{48}-\frac1{10}+\frac16=\frac{29}{480}\;,
\end{align*}$$
and if there’s a pattern, it isn’t an obvious one, but you can get as good an approximation as you want in relatively straightforward fashion;
$$F(x)=1+\frac{x}2-\frac{x^2}{12}+\frac{x^3}{24}-\frac{17x^4}{720}+\frac{29x^5}{480}+\dots$$
already gives two or three decimal places over much of the interval of convergence.
It leads to the same results. Call
$$
f(x) = \cos\sqrt{x}
$$
$$
f'(x) = -\frac{\sin\sqrt{x}}{2\sqrt{x}}
$$
So that
$$
\lim_{x\to 0 }f'(x) = -\frac{1}{2}
$$
where I used the well known limit $\lim_{t\to 0}\sin t/t = 1$.
$$
f''(x) = \frac{1}{4 x}\left(-\cos \sqrt{x} + \frac{\sin \sqrt{x}}{\sqrt{x}}\right)
$$
And the limit is
$$
\lim_{x\to0} f''(x) = \frac{1}{12}
$$
So the first two terms lead to
$$
\cos\sqrt{x} = 1 - \frac{x}{2} + \frac{x^2}{24} + \cdots\tag{1}
$$
For comparison, this is the expansion for $\cos (t)$
$$
\cos t = 1 -\frac{t^2}{2} + \frac{t^4}{24} + \cdots \tag{2}
$$
You can obtain (1) from (2) just by replacing $t = \sqrt{x}$
Best Answer
Hint
You can define a generalized expansion for $x$ near but $\neq 0$ by
$$\frac{1}{\sin(x)}=\frac{1}{x}\frac{1}{1-\frac{x^2}{6}+\frac{x^4}{5!}-...}$$
$$=\frac{1}{x}\left(1+\frac{x^2}{6}+\frac{7x^4}{360}+...\right).$$