Trignometric limit

The sum in the numerator evalutes to :

$= \dfrac{(2^{n}-1)(2^{n-1}-1)}{3}$

Proof :

We will be forming a recurrence relation and solving it :

$\displaystyle {S}_{n} = \displaystyle \sum_{r=1}^{{2}^{n-1}-1}{\tan^{2}{(\frac {r\pi}{{2}^{n}})}}$

It can also be written as :

$\displaystyle {S}_{n} = \displaystyle \sum_{r=1}^{{2}^{n-1}-1}{\cot^{2}{(\frac {r\pi}{{2}^{n}})}}$ (think why)

Adding these two forms we have :

$\displaystyle 2{S}_{n} = \sum _{ r=1 }^{ { 2 }^{ n-1 }-1 }{ \tan ^{ 2 }{ \frac { r\pi }{ { 2 }^{ n } } } +\cot ^{ 2 }{ \frac { r\pi }{ { 2 }^{ n } } } }$

Notice the identity $\tan ^{ 2 }{ \theta } +\cot ^{ 2 }{ \theta } =4\cot ^{ 2 }{ 2\theta } +2$

Using this we have :

$\displaystyle 2{S}_{n} = \sum _{ r=1 }^{ { 2 }^{ n-1 }-1 }{ 4\cot ^{ 2 }{ \frac { r\pi }{ { 2 }^{ n-1 } } } +2 }$

$\displaystyle \Rightarrow {S}_{n} = \sum _{ r=1 }^{ { 2 }^{ n-1 }-1 }{ (2\cot ^{ 2 }{ \frac { r\pi }{ { 2 }^{ n-1 } } } +1 )}$

Now notice that $\cot ^{ 2 }{ \frac { r\pi }{ { 2 }^{ n } } } =\cot ^{ 2 }{ \frac { ({ 2 }^{ n }-r)\pi }{ { 2 }^{ n } } }$

Using this identity we can rewrite our summation as :

$\displaystyle {S}_{n} = 4\sum _{ r=1 }^{ { 2 }^{ n-2 }-1 }{ \cot ^{ 2 }{ \frac { r\pi }{ { 2 }^{ n-1 } } } } +({ 2 }^{ n-1 }-1)$

We can also write it as :

$\displaystyle {S}_{n} = 4\sum _{ r=1 }^{ { 2 }^{ n-2 }-1 }{ \tan ^{ 2 }{ \frac { r\pi }{ { 2 }^{ n-1 } } } } +({ 2 }^{ n-1 }-1)$

Finally we have the recurrence relation as :

${ S }_{ r }=4{ S }_{ r-1 }+({ 2 }^{ r-1 }-1)$

To solve this recurrence relation I will use a small trick :

Let $(4^{r}){T}_{r} = {S}_{r}$

Our recurrence relation becomes :

${ T }_{ r }-{ T }_{ r-1 }=\frac { 1 }{ { 2 }^{ r+1 } } -\frac { 1 }{ { 4 }^{ r } }$

Summing both sides from $r=1$ to $r=n$ , we have :

${ T }_{ n }-{ T }_{ 1 }=\frac { 1 }{ 2 } (1-\frac { 1 }{ { 2 }^{ n } } )-\frac { 1 }{ 3 } (1-\frac { 1 }{ { 4 }^{ n } } )$

Observe that ${T}_{1}=0$

We have :

${ T }_{ n }=\frac { 1 }{ 2 } (1-\frac { 1 }{ { 2 }^{ n } } )-\frac { 1 }{ 3 } (1-\frac { 1 }{ { 4 }^{ n } } )$

Hence we have :

${ S }_{ n }=\frac { ({ 2 }^{ n }-1)({ 2 }^{ n-1 }-1) }{ 3 }$

The limit reaches a value of 3/4, therefore the answer is 3+4=7.