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.
 A: Hint:
$$ \frac{1}{n(n+1)(n+2)} = \frac{1/2}{n} - \frac{1}{n+1} + \frac{1/2}{n+2} $$
and
$$ 1 - \frac12 + \frac13 - \frac14 + \dotsb = \ln 2$$
A: You wish to find the sum $$\frac{1}{2} + \sum_{k=1}^{\infty} \frac{1}{(2k-1)(2k)(2k+1)}$$
Expanding the summand using partial fractions, we get
$$\frac{1}{(2k-1)(2k)(2k+1)}=\frac{A}{2k-1}+\frac{B}{2k}+\frac{C}{2k+1}$$$$ \implies 1=A(2k)(2k+1)+B(2k+1)(2k-1)+C(2k)(2k-1)$$
Solving this gives $A=C=\frac{1}{2},B=-1$. Thus splitting up our sum, we arrive at:
$$\frac{1}{2}+\frac{1}{2}\sum_{k=1}^{\infty}\frac{1}{2k-1}+\frac{1}{2}\sum_{k=1}^{\infty}\frac{1}{2k+1}-\sum_{k=1}^{\infty}\frac{1}{2k}$$
Now note that $$\sum_{k=1}^{\infty}\frac{1}{2k+1}=\sum_{k=1}^{\infty}\frac{1}{2k-1}-1$$
So our halves cancel, and grouping terms leaves us with:
$$\sum_{k=1}^{\infty}\frac{1}{2k-1}-\sum_{k=1}^{\infty}\frac{1}{2k}$$
In other words, $1-\frac{1}{2}+\frac{1}{3}-\ldots$ which is known to converge to $\ln(2)$
A: 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:
$$S = \frac{1}{1 - 1/2} = \boxed{2}.$$
