Summation and Factorials

I was feeling lazy to type the solution.

I would like to adjust the index to $\sum_{k=1}^{1998}\frac{k+2}{k!+(k+1)!+(k+2)!}$ . We first simplify the fraction to $\frac{k+2}{k!+(k+1)!+(k+2)!} = \frac{k+2}{k!(1+(k+1)+(k+1)(k+2))}=\frac{k+2}{k!(k+2)^2}$ From here, we can also put it into $\frac{1}{k!(k+2)}=\frac{(k+2)-1}{(k+2)!} = \frac{1}{(k+1)!}- \frac{1}{(k+2)!}$ Now we put it into sigma, which gives $\sum_{k=1}^{1998}(\frac{1}{(k+1)!}- \frac{1}{(k+2)!}) = \frac{1}{2!}-\frac{1}{2000!}\approx\frac{1}{2}$

$\sum _{ k=3 }^{ 2000 }{ \frac { k }{ (k-2)!+(k-1)!+k! } =\quad \frac { 1 }{ 3 } } +\frac { 1 }{ 8 } .....\\ Thus\quad the\quad value\quad of\quad this\quad summation\quad is\quad atleast\quad \frac { 1 }{ 3 } \\ Thereby\quad eliminating\quad the\quad options\quad less than\quad \frac { 1 }{ 3 } ,\quad the\quad only\quad possible\quad solution\quad is\quad \boxed { 0.5 } \\ Otherwise,\quad \sum _{ k=3 }^{ 2000 }{ \frac { k }{ (k-2)!+(k-1)!+k! } =\quad \quad } \sum _{ k=3 }^{ 2000 }{ \frac { k }{ (k-2)!({ k }^{ 2 }) } = } \sum _{ k=3 }^{ 2000 }{ \frac { 1 }{ (k-2)!({ k }) } = } \sum _{ k=3 }^{ 2000 }{ \frac { k-1 }{ (k)! } } \\ =\quad \sum _{ k=3 }^{ 2000 }{ \frac { 1 }{ (k-1)! } } -\quad \frac { 1 }{ k! } =\quad \left\{ \frac { 1 }{ 2! } -\frac { 1 }{ 2000! } \right\} \approx \boxed { 0.5 }$