Infinite series: $1/2 + 1/(1\cdot 2 \cdot 3) + 1/(3\cdot 4 \cdot 5) + \ldots$ How do I calculate this: $$\frac{1}{2}+\frac{1}{1\cdot 2\cdot 3}+\frac{1}{3\cdot 4\cdot 5}+\frac{1}{5\cdot 6\cdot 7}+\dots $$
I have not been sucessful to do this.
Hint:

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

and

$$ 1 - \frac12 + \frac13 - \frac14 + \dotsb = \ln 2$$
This is an infinite series, but it is not geometric because there is no common ratio.

So, let
$$S = 1 \cdot \frac{1}{2} + 2 \cdot \frac{1}{4} + 3 \cdot \frac{1}{8} + \dots + n \cdot \frac{1}{2^n} + \dotsb.$$
Multiplying this equation by 2, we get
$$2S = 1 \cdot 1 + 2 \cdot \frac{1}{2} + 3 \cdot \frac{1}{4} + \dots + n \cdot \frac{1}{2^{n - 1}} + \dotsb.$$
Subtracting these equations, we find
$$S = 1 \cdot 1 + 1 \cdot \frac{1}{2} + 1 \cdot \frac{1}{4} + \dots + 1 \cdot \frac{1}{2^n} + \dotsb.$$
So, $S=1+\dfrac12+\dfrac14+\dfrac18+\dotsb.$ Even though it didn't begin as one, we've managed to rewrite $S$ as an infinite geometric series. Thus, we may easily find its sum:
$$S = \frac{1}{1 - 1/2} = \boxed{2}.$$