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