Summing The Series

I will present you a manipulation mania.

$S=\displaystyle \sum _{ r=1 }^{ 2012 }{ \frac { (r+3)(r+2)(r+1)r }{ r!+(r+1)!+(r+2)! } }$

$S=\displaystyle \sum _{ r=1 }^{ 2012 }{ \frac { (r+3)(r+2)(r+1)r }{ { (r+2) }^{ 2 }r! } }$

$S=\displaystyle \sum _{ r=1 }^{ 2012 }{ \frac { (r+3){ (r+1) }^{ 2 }r }{ (r+2)! } }$

$S=\displaystyle \sum _{ r=1 }^{ 2012 }{ \frac { 1 }{ (r-2)! } +\frac { 3 }{ (r-1)! } -\frac { 1 }{ r! } +\frac { 2 }{ (r+1)! } -\frac { 2 }{ (r+2)! } }$

$S=\displaystyle \sum _{ r=1 }^{ 2012 }{ \frac { 1 }{ (r-2)! } +\sum _{ r=1 }^{ 2012 }{ \frac { 2 }{ (r-1)! } } + } \sum _{ r=1 }^{ 2012 }{ (\frac { 1 }{ (r-1)! } -\frac { 1 }{ r! } ) } +\sum _{ r=1 }^{ 2012 }{ (\frac { 2 }{ (r+1)! } -\frac { 2 }{ (r+2)! } ) }$

The last two summations are telescoping series and can be solved easily, and for the first two summations using the definition of euler's constant we can write :

$S=\displaystyle \sum _{ r=1 }^{ 2010 }{ \frac { 3 }{ r! } +\frac { 2 }{ 2011! } + } (1-\frac { 1 }{ 2012! } )+2(\frac { 1 }{ 2! } -\frac { 1 }{ 2014! } )$

Now $\displaystyle \sum _{ r=1 }^{ 2010 }{ \frac { 3 }{ r! } } \approx 3e$

Using this we get :

$S=(3e+2)+(\frac { 2 }{ 2011! } -\frac { 1 }{ 2012! } -\frac { 2 }{ 2014! } )$

Apart from $3e+2$ all the other terms are just small and can be neglected.

$S \approx 3e+2=10.15$

Hence $\left\lfloor S \right\rfloor =10$