Find sum of infinite anharmonic(?) series

I need help with this:

$$\frac{1}{1\cdot2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4\cdot5}+\frac{1}{3\cdot4\cdot5\cdot6}\dots$$ I don't know how to count sum of this series. It is similar to standard anharmonic series so it should have similar solution. However I can't work it out.

Answer

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

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