Function equal to infinite series $\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n+3)(2n+1)!}$ I'd like to know if there is a simple function equivalent of $$\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n+3)(2n+1)!}$$
I recognize that it looks similar to $\frac{\sin{x}}{x}$, but with an extra $(2n+3)$ factor in the denominator that I don't know how to account for. I've graphed the sum from $n=0$ to $100$ on an online graphing calculator and tried to find a function that matches it, but I've had no luck. I noticed that $\frac{\sin{x}}{3x}$ looks similar, but it is a very naive guess and I don't have a better one. It may even be the case that such a function does not exist.

If you had a $2n+2$ in the denominator, it would look like $$g(x) = \sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n+3)!}. \tag{1}$$  If you multiplied $g(x)$ by $x^2$ and took the derivative,
$$\begin{align}
\frac{d}{dx}[x^2 g(x)] &= \sum_{n=0}^\infty (-1)^n \frac{d}{dx}\left[ \frac{x^{2n+2}}{(2n+3)!} \right] \\
&= \sum_{n=0}^\infty (-1)^n \frac{(2n+2) x^{2n+1}}{(2n+3)(2n+2)(2n+1)!} \\
&= \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{(2n+3)(2n+1)!}. \tag{2}\end{align}$$
This gets you very close to what you want; all you need to do is factor out one power of $x$:
$$f(x) = \sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n+3)(2n+1)!} = \frac{1}{x} \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{(2n+3)(2n+1)!} = \frac{1}{x} \frac{d}{dx}\left[ x^2 g(x) \right] \tag{3}$$ where $g$ is defined as in $(1)$.  So what is $g$?  We know that $$\sin x = \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{(2n+1)!}. \tag{4}$$  So $$\sin x - x = \sum_{n=1}^\infty (-1)^n \frac{x^{2n+1}}{(2n+1)!} = \sum_{n=0}^\infty (-1)^{n+1} \frac{x^{2n+3}}{(2n+3)!}. \tag{5}$$It follows that $$g(x) = \frac{x - \sin x}{x^3}, \tag{6}$$ from which we use $(3)$ to compute $f(x)$.
