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.
 A: 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 $f(x)$.
A: $$f(x)=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n+3)(2n+1)!}$$
$$x^3f(x)=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+3}}{(2n+3)(2n+1)!}$$
$$\big[x^3f(x)\big]'=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+2}}{(2n+1)!}=x\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!}=x \sin(x)$$
$$x^3f(x)=\int x\sin(x)\,dx=\sin (x)-x \cos (x)$$
$$f(x)=\frac{\sin (x)-x \cos (x) }{x^3}$$
A: This is to follow through @user170231's hint. As he noted, the sum boils down to computing the two sums
$$
S_1(x)=\sum_{n = 0}^{\infty} \frac{(-1)^n x^{2n}}{(2n+1)!} \quad\text{and}\quad S_2(x)=\sum_{n = 0}^{\infty} \frac{1}{2n + 3}\frac{(-1)^n x^{2n}}{(2n)!}
$$
For the first sum it is quite easy. We simply recall the trigonometric series
$$
\cos x = \sum_{n = 0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} \quad\text{and}\quad \sin x = \sum_{n = 0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}
$$
Then it is immediately clear that $S_1(x) = \frac{\sin x}{x}$. I wrote out both series because we will need it now for the second sum, and it will get annoying real quick! The idea is to keep integrating $S_2(x)$ so that the power of $x$ increases, until the denominator "merges" with the $2n + 3$, from which we will apply some trigonometric series. Also, we can safely ignore constants since we will be taking derivatives later, though I encourage you to work them out by looking at special values.
Integrating once:
$$
\int S_2(x) dx = \sum_{n = 0}^{\infty} \frac{1}{2n + 3} \frac{(-1)^n x^{2n + 1}}{(2n + 1)!}
$$
Integrating twice (this isn't really a double integral so I don't write $\iint$):
$$
\int\left(\int S_2(x) dx\right)dx = \sum_{n = 0}^{\infty} \frac{1}{2n + 3} \frac{(-1)^n x^{2n + 2}}{(2n + 2)!} = \sum_{n = 0}^{\infty} \frac{(-1)^n x^{2n + 2}}{(2n + 3)!}
$$
As we can see, $\int\left(\int S_2(x) dx\right)dx$ is just a shifted version of $\frac{\sin x}{x}$. In fact by changing some variables, you just get
$$
\sum_{n = 0}^{\infty} \frac{(-1)^n x^{2n + 2}}{(2n + 3)!} = \frac{1}{x} \sum_{n = 1}^{\infty} \frac{(-1)^{n + 1} x^{2n + 1}}{(2n + 1)!} = -\frac{1}{x} \left(\sin x - x\right) = 1 - \frac{\sin x}{x}
$$
Therefore, we can recover $S_2(x)$ by taking derivatives:
\begin{align*}
S_2(x) &= -\frac{d}{dx} \frac{x\cos x - \sin x}{x^2} \\
&= -\frac{(\cos x - x\sin x - \cos x)x^2 - 2x(x\cos x - \sin x)}{x^4} \\
&= -\frac{-x^3\sin x - 2x^2\cos x + 2x\sin x}{x^4} \\
&= \frac{x^2\sin x + 2x\cos x - 2\sin x}{x^3}
\end{align*}
Then the sum is just ading them up and rescaling according to partial fractions.
