Finding the infinite Sum of a series: $\sum\frac1{n(n+1)(n+2)}$

Question

Find the infinite Sum of the series with general term $\frac{1}{n(n+1)(n+2)}$.

I decomposed the fraction upto this $1/(2n)-1/(n+1)+1/(2n+4)$. But I find no link about cancelling terms. So how should I find the infinite Sum?

Please don't say that this is off topic or homework. if wanna say that this is homework, at least give a hint if I am going the right way?

Answer 1

Hint:

$$\frac1{n(n+2)}=\frac12\left(\frac1n-\frac1{n+2}\right)$$

$$\frac1{n(n+1)(n+2)}=\frac12\left(\frac1{n(n+1)}-\frac1{(n+1)(n+2)}\right)$$

Answer 2

Hint. You may write $$ \frac1{n(n+1)(n+2)}=\frac12\left(\frac1n-\frac1{n+1}\right)-\frac12\left(\frac1{n+1}-\frac1{n+2}\right). $$ and see that terms telescope.

Answer 3

Alongside Olivier's approach, you can make a good old table of values. Make a table long enough of $n$ (I assume starting $n=1$) This is how they cancel; you got to look diagonally: $\frac{1}{6}$ - $\frac{1}{3}$ + $\frac{1}{6}$ followed by $\frac{1}{8}$ - $\frac{1}{4}$ + $\frac{1}{8}$ followed by $\frac{1}{10}$ - $\frac{1}{5}$ + $\frac{1}{10}$ etc. Surviving terms are $\frac{1}{2}$ , -$\frac{1}{2}$ , $\frac{1}{4}$ from which you can see the answer.

Answer 4

Notice, use partial fractions in form of difference of two terms as follows $$\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)}$$$$=\sum_{n=1}^{\infty}\frac 12\left(\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}\right)$$ $$=\small \frac 12\lim_{n\to \infty}\left(\left(\frac{1}{1\cdot 2}-\frac{1}{2\cdot 3}\right)+\left(\frac{1}{2\cdot 3}-\frac{1}{3\cdot 4}\right)+\left(\frac{1}{3\cdot 4}-\frac{1}{4\cdot 5}\right)+\ldots+\left(\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}\right)\right)$$ $$=\frac 12\lim_{n\to \infty}\left(\frac{1}{1\cdot 2}-\frac{1}{(n+1)(n+2)}\right)$$ $$=\frac 12\left(\frac{1}{2}-0\right)=\color{red}{\frac 14}$$