Let S _ k = ∑ _ r = 1 ^ k ^ - 1 ( 6 ^ r 2 ^ 2r + 1 + 3 ^ 2r + 1 ),...

Given ${S}_{k} = \mathop{∑}_{r = 1}^{k}\tan^{- 1}\left(\dfrac{{6}^{r}}{{2}^{2r + 1} + {3}^{2r + 1}}\right)$

$⇒ {S}_{k} = \mathop{∑}_{r = 1}^{k}\tan^{- 1}\left(\dfrac{{2}^{r} \cdot {3}^{r}}{{2}^{2r} \cdot 2 + {3}^{2r} \cdot 3}\right)$

Divide by ${3}^{2r}$ we get,

${S}_{k} = \mathop{∑}_{r = 1}^{k}\tan^{- 1}\left(\dfrac{{\left(\dfrac{2}{3}\right)}^{r}}{{\left(\dfrac{2}{3}\right)}^{2r} . 2 + 3}\right)$

${S}_{k} = \mathop{∑}_{r = 1}^{k}\tan^{- 1}\left(\dfrac{{\left(\dfrac{2}{3}\right)}^{r}}{3\left({\left(\dfrac{2}{3}\right)}^{2r + 1} + 1\right)}\right)$

Let ${\left(\dfrac{2}{3}\right)}^{r} = t$

${S}_{k} = \mathop{∑}_{r = 1}^{k}\tan^{- 1}\left(\dfrac{\dfrac{t}{3}}{1 + \dfrac{2}{3}{t}^{2}}\right)$

${S}_{k} = \mathop{∑}_{r = 1}^{k}\tan^{- 1}\left(\dfrac{t - \dfrac{2t}{3}}{1 + t . \dfrac{2t}{3}}\right)$

${S}_{k} = \mathop{∑}_{r = 1}^{k}\left(\tan^{- 1}\left(t\right) - \tan^{- 1}\left(\dfrac{2t}{3}\right)\right)$

${S}_{k} = \mathop{∑}_{r = 1}^{k}\left(\tan^{- 1}{\left(\dfrac{2}{3}\right)}^{r} - \tan^{- 1}{\left(\dfrac{2}{3}\right)}^{r + 1}\right)$

${S}_{k} = \left(\tan^{- 1}\left(\dfrac{2}{3}\right) - \tan^{- 1}{\left(\dfrac{2}{3}\right)}^{2}\right) + \left(\tan^{- 1}{\left(\dfrac{2}{3}\right)}^{2} - \tan^{- 1}{\left(\dfrac{2}{3}\right)}^{3}\right) + . . . + \left(\tan^{- 1}{\left(\dfrac{2}{3}\right)}^{k} - \tan^{- 1}{\left(\dfrac{2}{3}\right)}^{k + 1}\right)$

${S}_{k} = \tan^{- 1}\left(\dfrac{2}{3}\right) - \tan^{- 1}{\left(\dfrac{2}{3}\right)}^{k + 1}$

Then, ${S}_{\infty} = \lim_{k \to \infty}\left(\tan^{- 1}\left(\dfrac{2}{3}\right) - \tan^{- 1}{\left(\dfrac{2}{3}\right)}^{k + 1}\right)$

$= \tan^{- 1}\left(\dfrac{2}{3}\right) - \tan^{- 1}\left(0\right)$

$∴ {S}_{\infty} = \tan^{- 1}\left(\dfrac{2}{3}\right) = \cot^{- 1}\left(\dfrac{3}{2}\right), \left(∵ \tan^{- 1}x = \cot^{- 1}\left(\dfrac{1}{x}\right)\right)$