<s>[INST] Prove $\sin x + \arcsin x > 2x$ using Maclaurin series My teacher asked us to solve this problem using the Maclaurin series, but I could not figure out how to approach..
Prove that the inequality sin x + arcsin x > 2x holds for all values of x such
that 0 < x ≤ 1.
I know that the Maclaurin series of 
sin(x) = x - $\frac{x^3}{3!}$ + $\frac{x^5}{5!}$ - $\frac{x^7}{7!}$ + ...
arcsin(x) = x + $\frac{1}{2}\cdot\frac{x^3}{3}$ + ($\frac{1}{2}\cdot\frac{3}{4}$)$\cdot\frac{x^5}{5}$ + ...
However, I do not know how to prove this using there series...Could anyone have some ideas?
Thank you!
 [/INST] For $0 < x \le 1$ we have
$$
 \sin(x) + \arcsin(x) = 2 x + \sum_{n=2}^\infty \left(\frac{(-1)^n}{(2n+1)!} +  \frac{1 \cdot 3 \cdots (2n-1)}{2 \cdot 4 \cdots (2n)} \frac{1}{2n+1} \right) x^{2n+1}
$$
because the $x^3$ terms cancel. Therefore it suffices to show that
$$
 \frac{1 \cdot 3 \cdots (2n-1)}{2 \cdot 4 \cdots (2n)} \frac{1}{2n+1} > \frac{1}{(2n+1)!}
$$
for $n \ge 2$. This is equivalent to
$$
\frac{1 \cdot 3 \cdots (2n-1)}{2 \cdot 4 \cdots (2n)} \cdot (2n)! > 1
$$
or 
$$
\bigl( 1 \cdot 3 \cdots (2n-1) \bigr)^2 > 1 \, ,
$$
which is obvious.
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