One of my homework problems was to determine whether the series $\sum_{n=0}^\infty \frac{(-1)^n x^{2n-4}}{(2n-1)!}$ converges, and if it does, find what it converges to.
I used the ratio test to show that it converges absolutely. To find what it converges to, I first multiplied it by $x^3$:
$f(x)=\sum_{n=0}^\infty \frac{(-1)^n x^{2n-4}}{(2n-1)!} \implies x^3f(x)=\sum_{n=0}^\infty \frac{(-1)^n x^{2n-1}}{(2n-1)!}$.
I know that this is almost the Taylor series of $\sin x$, but I couldn't remember whether or not we proved this in class. So, I said:
"By Taylor's formula, we have:
$\sin x = x - \frac{x^3}{3!}+\frac{x^5}{5!}-\dots + \frac{(-1)^{n+1}x^{2n-1}}{(2n-1)!} + R_n(x)$, where $R_n(x)$ is the error term.
$R_n(x)=\frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}$.
For all $c$, we have $f^{(n+1)}(c) \leq 1$. This means that:
$R_n(x) \leq \frac{x^{n+1}}{(n+1)!}$.
Hence, by the Remainder Estimation Theorem, the series converges to $f(x)$ for all $x$."
I then concluded that the original series must converge to $-\frac{\sin x}{x^3}$.
Is my solution correct? Also, are my arguments easy to read or is it confusing?
I apologize if this is a trivial question, but I don't have anyone to ask this and I want to make sure I understand what I am doing.