Limit of a sequence : $x_n = \frac{1}{1\cdot 2\cdot 3} + \frac{1}{2\cdot 3\cdot 4} + \ldots + \frac{1}{n(n+1)(n+2)}$

Question

Can someone help me with this problem?

Finding the limit $\lim_{n \to \infty}\ x_n$ where

$$x_n = \frac{1}{1\cdot 2\cdot 3} + \frac{1}{2\cdot 3\cdot 4} + \ldots + \frac{1}{n(n+1)(n+2)},\quad n\in\mathbb{N}.$$ I don't have a clue how to do this.

Answer 1

HINT: use that $$\frac{1}{i(i+1)(i+2)}=1/2\, \left( i+2 \right) ^{-1}- \left( i+1 \right) ^{-1}+1/2\,{i}^{-1}$$

Answer 2

Hint. On may observe that $$ \frac{1}{n(n+1)(n+2)}=\frac12\frac{2}{n(n+1)(n+2)}=\frac12\frac{(n+2)-n}{n(n+1)(n+2)} $$ giving $$ \frac{1}{n(n+1)(n+2)}=\frac{1}{2n(n+1)}-\frac{1}{2(n+1)(n+2)} $$ the given sum is thus a telescoping one.

Hope you can finish it.

Answer 3

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

Answer 4

From the question here we find that $$\sum_{r=1}^\infty \frac 1{r^\overline{m}}=\frac 1{(m-1)(m-1)!}$$ Putting $m=3$ gives

$$\sum_{r=1}^\infty \frac 1{r^\overline{3}}=\sum_{r=1}^\infty \frac 1{r(r+1)(r+2)}=\frac 1{2\cdot 2!}=\color{red}{\frac 14}$$