Find sum of infinite anharmonic(?) series

What you're evaluating is the infinite series

$$\sum_{n = 1}^\infty \frac{1}{n(n+1)(n+2)(n+3)}.$$

Since

$$\frac{1}{n(n+3)} = \frac{1}{3n}-\frac{1}{3(n+3)},$$

then

$$\frac{1}{n(n+1)(n+2)(n+3)} = \frac{1}{3n(n+1)(n+2)}-\frac{1}{3(n+1)(n+2)(n+3)}.$$

So your series telescopes to

$$\frac{1}{3(1)(2)(3)} = \frac{1}{18}.$$

$$I=\dfrac2{n(n+1)(n+2)(n+3)}$$

$$2I=\dfrac{(n+2)(n+1)-n(n+3)}{n(n+1)(n+2)(n+3)}=\dfrac1{n(n+3)}-\dfrac1{(n+1)(n+2)}$$

$$6I=\dfrac1n-\dfrac1{n+3}-3\left(\dfrac1{n+1}-\dfrac1{n+2}\right)$$

Can you see the telescoping nature?

Using Euler's Beta function: $$\begin{eqnarray*}\sum_{k=1}^{+\infty}\frac{1}{k(k+1)(k+2)(k+3)}&=&\sum_{k\geq 1}\frac{\Gamma(k)}{\Gamma(k+4)}=\frac{1}{\Gamma(4)}\sum_{k\geq 1}B(k,4)\\&=&\frac{1}{6}\sum_{k\geq 1}\int_{0}^{1}(1-x)^3 x^{k-1}\,dx\\&=&\frac{1}{6}\int_{0}^{1}(1-x)^2\,dx\\&=&\frac{1}{6}\int_{0}^{1}x^2\,dx = \color{red}{\frac{1}{18}}.\end{eqnarray*}$$ Using creative (not so much) telescoping, for $a_k=\frac{1}{k(k+1)(k+2)}$ we have $a_{k}-a_{k+1}=\frac{3}{k(k+1)(k+2)(k+3)}$, so $$\sum_{k\geq 1}\frac{1}{k(k+1)(k+2)(k+3)}=\frac{a_1}{3}=\frac{1}{3\cdot 3!}=\frac{1}{18}.$$