<s>[INST] Is there a closed-form of $ \sum_{n=0}^{\infty }\frac{(-1)^n}{(2n+1)(2n+2)(2n+3)(2n+4)}$ Is there a closed-form of $$\sum_{n=0}^{\infty }\frac{(-1)^n}{(2n+1)(2n+2)(2n+3)(2n+4)}$$
Thanks for any help
 [/INST] Let $$f(x)=\sum_{n=0}^{\infty }\frac{(-1)^nx^{2n+4}}{(2n+1)(2n+2)(2n+3)(2n+4)}.$$
Then taking the derivatives wrt $x$, the factors in the denominator disappear:
$$f''''(x)=\sum_{n=0}^{\infty }(-x^2)^n=\frac1{1+x^2}.$$
Integrate four times from $0$ to $x$ (factors added for convenience).
$$\begin{align}f'''(x)&=\sum_{n=0}^{\infty }\frac{(-1)^nx^{2n+1}}{2n+1}\\
&=\arctan x.\end{align}$$
$$\begin{align}2f''(x)&=2\sum_{n=0}^{\infty }\frac{(-1)^nx^{2n+2}}{(2n+1)(2n+2)}\\
&=2x\arctan x-\log(x^2+1).\end{align}$$
$$\begin{align}2f'(x)&=2\sum_{n=0}^{\infty }\frac{(-1)^nx^{2n+3}}{(2n+1)(2n+2)(2n+3)}\\
&=(x^2-1)\arctan x-x\log(x^2+1)+x.\end{align}$$
$$\begin{align}12f(x)&=12\sum_{n=0}^{\infty }\frac{(-1)^nx^{2n+4}}{(2n+1)(2n+2)(2n+3)(2n+4)}\\
&=2(x^2-3)x\arctan x-(3x^2-1)\log(x^2+1)+5x^2.\end{align}$$
Then
$$f(1)=\dfrac{-\pi-2\log2+5}{12}.$$
For the same "price", you also get the sums with $1$ to $3$ factors: $f'''(1)=\pi/4$, $f''(1)=\pi/4-\log2/2$, $f'(1)=-\log2/2+1/2$.

For efficient evaluation of the integrals, we can proceed as follows using complex numbers.
Notice that $\dfrac1{x^2+1}$ is the imaginary part of $\dfrac1{x-i}$, and use the change of variable $t=x-i$.
Integrate four times (by parts) between $-i$ and $t$:
$$\frac1t,$$
$$\log t,$$
$$t(\log t-1),$$
$$\frac{t^2}4(2\log t-3),$$
$$\frac{t^3}{36}(6\log t-11).$$
Then take the imaginary parts of the differences of the antiderivative values between $1-i$ and $-i$:
$$\Im\left(\log(1-i)-\log(-i)\right)=\frac\pi4,$$
$$\Im\left((1-i)\left(\log(1-i)-1\right)-(-i)\left(\log(-i)-1\right)\right)=-\frac\pi4-\frac{\log2}2,$$
$$\Im\left(\frac{(1-i)^2}4(2\log(1-i)-3)-\frac{(-i)^2}4(2\log(-i)-3)\right)=\frac32-\pi-\frac{\ln2}2$$
$$\Im\left(\frac{(1-i)^3}{36}\left(6\log(1-i)-11\right)-\frac{(-i)^3}{36}\left(6\log(-i)-11\right)\right)=\frac\pi{12}-\frac{\log2}6-\frac{11}{12}$$
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