Finding $\frac{3}{1!+2!+3!}+\frac{4}{2!+3!+4!} +\frac{5}{3!+4!+5!}+\cdots+\frac{2008}{2006!+2007!+2008!}$ [duplicate]

Solution 1:

What we want is $$\sum_{n=1}^{N} \dfrac{n+2}{n! + (n+1)! +(n+2)!}$$ \begin{align} \dfrac{n+2}{n! + (n+1)! +(n+2)!} & = \dfrac{n+2}{n! \left( 1 + (n+1) + (n+1)(n+2) \right)}\\ & = \dfrac{n+2}{n! \left( n^2 + 4n + 4 \right)}\\ & = \dfrac1{n! \left( n+2 \right)}\\ & = \dfrac{n+1}{(n+2)!}\\ & = \dfrac{n+2}{(n+2)!} - \dfrac1{(n+2)!}\\ & = \dfrac1{(n+1)!} - \dfrac1{(n+2)!} \end{align}

Can you finish it off from here?

Move your mouse over the gray area below for the complete answer.

\begin{align}\sum_{n=1}^{N} \dfrac{n+2}{n! + (n+1)! +(n+2)!} & = \sum_{n=1}^{N} \left( \dfrac1{(n+1)!} - \dfrac1{(n+2)!}\right)\\ & = \left( \dfrac1{2!} - \dfrac1{3!} + \dfrac1{3!} - \dfrac1{4!} + \dfrac1{4!} - \dfrac1{5!} + \cdots + \dfrac1{(N+1)!} - \dfrac1{(N+2)!}\right)\\ & = \dfrac1{2!} - \dfrac1{(N+2)!}\end{align} Set $N=2006$ to get the answer to your question.

Solution 2:

Hint: Apply telescoping series to $$ \begin{align} \sum_{k=1}^{2006}\frac{k+2}{k!+(k+1)!+(k+2)!} &=\sum_{k=1}^{2006}\frac{k+2}{k!\,(k+2)^2}\\ &=\sum_{k=1}^{2006}\frac{k+1}{(k+2)!}\\ &=\sum_{k=1}^{2006}\left(\frac1{(k+1)!}-\frac1{(k+2)!}\right) \end{align} $$

Solution 3:

$$\dfrac1{n(n-2)!}=\dfrac{n-1}{n!}$$ which is Telescoping