Infinite Sum

We have $S=\sum _{ r=0 }^{ n }{ \frac { { 2 }^{ r } }{ { 2 }^{ { 2 }^{ r } }+1 } }$

Hence ${T}_{r}=\frac { { 2 }^{ r } }{ { 2 }^{ { 2 }^{ r } }+1 }$

So ${T}_{r}$ can be written as

$=(\frac { { 2 }^{ r } }{ { 2 }^{ { 2 }^{ r } }-1 } )-(\frac { { 2 }^{ r+1 } }{ { 2 }^{ { 2 }^{ r+1 } }-1 } )=V(r)-V(r+1)$

So we have formed a telescoping series. Finally $S$ can be written as :

$S=V(0)-V(n+1)=1-(\frac { { 2 }^{ n+1 } }{ { 2 }^{ { 2 }^{ n+1 } }-1 } )$

Finally our infinte series sum can be written as :

$S=1-\lim _{ n\rightarrow \infty }{ (\frac { { 2 }^{ n+1 } }{ { 2 }^{ { 2 }^{ n+1 } }-1 } ) }$

$\Rightarrow \boxed { S=1 }$