Let $\mathrm{S_{k}}=\sum_{\mathrm{r}=1}^{\mathrm{k}} \tan ^{-1}\left(\frac{6^{\mathrm{r}}}{2^{2 \mathrm{r}+1}+3^{2 \mathrm{r}+1}}\right)$. Then $\lim _{\mathrm{k} \rightarrow \infty} \mathrm{S_{k}}$ is equal to :

(A) $\cot ^{-1}\left(\frac{3}{2}\right)$

(B) $\tan ^{-1}\left(\frac{3}{2}\right)$

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Inverse Trigonometric Functions

Let

S_{k} = \sum_{r = 1}^{k} tan^{- 1} (\frac{6 ^{r}}{2 ^{2 r + 1} + 3 ^{2 r + 1}})

. Then

lim_{k \to \infty} S_{k}

is equal to :

Answer: (A)

Put

x_{r} = (\frac{3}{2})^{r}

. Then

\frac{x _{r + 1} - x _{r}}{1 + x _{r} x _{r + 1}} = \frac{\frac{1}{2} ( \frac{3}{2} ) ^{r}}{1 + ( \frac{3}{2} ) ^{2 r + 1}} = \frac{3 ^{r}}{2 ^{r + 1}} \cdot \frac{2 ^{2 r + 1}}{2 ^{2 r + 1} + 3 ^{2 r + 1}} = \frac{6 ^{r}}{2 ^{2 r + 1} + 3 ^{2 r + 1}}

and since

x_{r} x_{r + 1} > 0

, each term of the sum is

tan^{- 1} x_{r + 1} - tan^{- 1} x_{r}

. The sum therefore telescopes:

S_{k} = tan^{- 1} (\frac{3}{2})^{k + 1} - tan^{- 1} \frac{3}{2}

k \to \infty

the first term tends to

\frac{π}{2}

, so

k \to \infty lim S_{k} = \frac{π}{2} - tan^{- 1} \frac{3}{2} = cot^{- 1} \frac{3}{2}

The limit is

cot^{- 1} \frac{3}{2}

JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Single correct option