I am boxing sequences

$\displaystyle{S=\sum _{ k=1 }^{ 100 }{ \cfrac { 1 }{ { x }_{ k }+1 } } =\sum _{ k=1 }^{ 100 }{ \cfrac { { x }_{ k } }{ { x }_{ k }({ x }_{ k }+1) } } \\ \because \begin{cases} { x }_{ k }({ x }_{ k }+1)={ x }_{ k+1 } \\ { { x }_{ k } }^{ 2 }={ x }_{ k+1 }-{ x }_{ k } \end{cases}\\ S=\sum _{ k=1 }^{ 100 }{ \cfrac { { x }_{ k } }{ { x }_{ k+1 } } } =\sum _{ k=1 }^{ 100 }{ \cfrac { { { x }_{ k } }^{ 2 } }{ { (x }_{ k+1 }){ x }_{ k } } } \\ S=\sum _{ k=1 }^{ 100 }{ \cfrac { { x }_{ k+1 }-{ x }_{ k } }{ { (x }_{ k+1 }){ x }_{ k } } } =\sum _{ k=1 }^{ 100 }{ \cfrac { 1 }{ { x }_{ k } } -\cfrac { 1 }{ { x }_{ k+1 } } } \\ \left\{ \because Telescopic \right\} \\ S=\cfrac { 1 }{ { x }_{ 1 } } -\cfrac { 1 }{ { x }_{ 100 } } =2-\cfrac { 1 }{ { x }_{ 100 } } \\ }$

Now we don't need to evaluate ${ x }_{ 100 }$ But we know that given sequence is Increasing , Since It's Nature is Parabolic, So it's increses rapidly , and clearly , $\displaystyle{{ x }_{ 3 }>1\Rightarrow { x }_{ 100 }>>>1\Rightarrow \cfrac { 1 }{ { x }_{ 100 } } <<<1\\ 1^{ - }<S<{ 2 }^{ - }\\ \boxed { \left\lfloor S \right\rfloor =1 } }$