How to solve $\frac 1{1.2.3}+\frac 1{2.3.4}+\frac 1{3.4.5}+...$
As my question says,
$$\frac 1{1.2.3}+\frac 1{2.3.4}+\frac 1{3.4.5}+...$$
As far as $\frac 1{1.2}+\frac 1{2.3}+\frac 1{3.4}+...$ is concerned, we can write it as $\frac 1{n-1} - \frac 1{n}$ What can we do here?
Solution 1:
We can write it as $\frac 1{n(n+1)} - \frac 1{(n+1)(n+2)}$ with a factor of $0.5$
Solution 2:
Write the sum as $$\sum_{n\ge 2}\frac{1}{n(n^2-1)}=\frac{1}{2}\sum_{n\ge 2}\left(\frac{1}{n-1}+\frac{1}{n+1}-\frac{2}{n}\right)=\frac{1}{4}$$