A sum of series problem: $\frac{3}{1!+2!+3!} + \frac{4}{2!+3!+4!} + \cdots + \frac{2008}{2006!+2007!+2008!}$

Question

I have a question regarding the sum of this series: $$\frac{3}{1!+2!+3!} + \frac{4}{2!+3!+4!} + \cdots + \frac{2008}{2006!+2007!+2008!}$$ My approach: I found that this sum is equal to: $$\sum_{n=3}^{2008}\frac{n}{(n-2)!+(n-1)!+(n)!}$$

I reduced it to : $$\sum_{n=3}^{2008}\frac{1}{n(n-2)!}$$
Please suggest how to proceed further.

Answer 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?

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\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.

Answer 2

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

Answer 3

$\sum_{n=3}^{2008}\frac{1}{n(n-2)!}=\sum_{n=3}^{2008}\frac{1}{n(n-2)!}.\frac{(n-1)}{(n-1)}=\sum_{n=3}^{2008}\frac{n-1}{n!}=\sum_{n=3}^{2008}(\frac{n}{n!}-\frac{1}{n!})=\sum_{n=3}^{2008}(\frac{1}{(n-1)!}-\frac{1}{n!})=\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...-\frac{1}{2007!}+\frac{1}{2007!}-\frac{1}{2008!}=\frac{1}{2!}-\frac{1}{2008!}$

Answer 4

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