Which is greater ? Sum of odd power terms or even power terms in the exponential Taylor series?

Question

I came across this question, in a book.

Define $f(x) = \sum_{n=0}^{\infty} \frac{x^{2n+1}}{{(2n+1)}!} $ and $ g(x) = \sum_{n=0}^{\infty} \frac{x^{2n}}{{(2n)}!} $, where x is a real number. Then, which of the following statements is correct?

(A) $f(x) > g(x)$ for all x

(B) $f(x) < g(x)$ for all x

(C) $f(x) = g(x)$ for all x

(D) none of the above statements need necessarily hold for all x

I tried to write out the Taylor series for $e^{-x}$ and used the fact that the series would still be positive and hence concluded that the sum of even powered terms must be greater than the odd powered terms, id est, statement (B).

However, the answer key given in the book says it should be option (A). Is there any thing wrong with my approach?

Answer 1

We have $2f(x)=e^x-e^{-x}$ and $2g(x)=e^x+e^{-x}$, so $g(x)\gt f(x)$.

Answer 2

A can not possibly be correct because, for negative x, f(x) < 0 and g(x) > 0.

My choice is D, but I don't have a proof right now.