The sum of the infinite series \cot ^-1(\frac74)+\cot ^-1(\frac194)+\cot ^-1(\frac394)+\cot ^-1(\frac674)+\ldots . is :

Inverse Trigonometric Functions PYQ — JEE Main 2025 · 4 Apr · Shift 2 · Q40 with Solution

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Correct answer: C

Identify the pattern of the general term

The given series is

\cot^{-1}\left(\frac{7}{4}\right)+\cot^{-1}\left(\frac{19}{4}\right)+\cot^{-1}\left(\frac{39}{4}\right)+\cot^{-1}\left(\frac{67}{4}\right)+\cdots

Look at the numerators: $7,\ 19,\ 39,\ 67,\ldots$ These are of the form $n^2+(n+1)^2$ for $n=1,2,3,4,\ldots$ , because $1^2+2^2=5\quad(\text{not }7),$ so let us check another form: $2n^2+2n+3$ For $n=1,2,3,4$ : $2(1)^2+2(1)+3=7$ $2(2)^2+2(2)+3=15\neq 19$ So try $2n^2+4n+1$ For $n=1,2,3,4$ : $7,17,31,49\neq$

Now observe the successive differences: $19-7=12,\quad 39-19=20,\quad 67-39=28$ Second difference is constant $8$ , so numerator is quadratic: $an^2+bn+c$ with $2a=8\Rightarrow a=4$ .

Let numerator be $4n^2+bn+c$ Using $n=1,2$ : $4+b+c=7 \Rightarrow b+c=3$ $16+2b+c=19 \Rightarrow 2b+c=3$ Subtracting gives $b=0$ , hence $c=3$ .

So the $n$ th term is $\cot^{-1}\left(\frac{4n^2+3}{4}\right).$

Thus $S=\sum_{n=1}^{\infty} \cot^{-1}\left(\frac{4n^2+3}{4}\right).$

Convert each term into a telescoping form

We use the identity $\cot^{-1}x-\cot^{-1}y=\cot^{-1}\left(\frac{xy+1}{y-x}\right)$ for suitable positive values.

Now compute $\cot^{-1}(2n-1)-\cot^{-1}(2n+1).$ Then

=\frac{4n^2-1+1}{2} =\frac{4n^2}{2}=2n^2.$$ That does not match. Instead try $$\cot^{-1}\left(\frac{2n-1}{2}\right)-\cot^{-1}\left(\frac{2n+1}{2}\right).$$ Using the same identity: $$x=\frac{2n-1}{2},\quad y=\frac{2n+1}{2}$$ Then $$xy+1=\frac{(2n-1)(2n+1)}{4}+1=\frac{4n^2-1}{4}+1=\frac{4n^2+3}{4}$$ and $$y-x=\frac{2n+1}{2}-\frac{2n-1}{2}=1.$$ Therefore $$\cot^{-1}\left(\frac{2n-1}{2}\right)-\cot^{-1}\left(\frac{2n+1}{2}\right) =\cot^{-1}\left(\frac{4n^2+3}{4}\right).$$ So each term becomes $$\cot^{-1}\left(\frac{4n^2+3}{4}\right)=\cot^{-1}\left(\frac{2n-1}{2}\right)-\cot^{-1}\left(\frac{2n+1}{2}\right).$$ --- 3. **Write the series as a telescoping sum** Hence

S=\sum_{n=1}^{\infty}\left[\cot^{-1}\left(\frac{2n-1}{2}\right)-\cot^{-1}\left(\frac{2n+1}{2}\right)\right].

Expanding,

S=\left[\cot^{-1}\left(\frac{1}{2}\right)-\cot^{-1}\left(\frac{3}{2}\right)\right] +\left[\cot^{-1}\left(\frac{3}{2}\right)-\cot^{-1}\left(\frac{5}{2}\right)\right] +\cdots

All middle terms cancel, so

S=\cot^{-1}\left(\frac{1}{2}\right)-\lim_{n\to\infty}\cot^{-1}\left(\frac{2n+1}{2}\right).

Since $$\lim_{n\to\infty}\cot^{-1}\left(\frac{2n+1}{2}\right)=0,$$ we get $$S=\cot^{-1}\left(\frac{1}{2}\right).$$ --- 4. **Match with the options** Use the identity for positive $x$: $$\cot^{-1}x=\frac{\pi}{2}-\tan^{-1}x.$$ Thus

\cot^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{2}-\tan^{-1}\left(\frac{1}{2}\right).

So the sum is $$\boxed{\frac{\pi}{2}-\tan^{-1}\left(\frac{1}{2}\right)}.$$ This is **Option C**. --- 5. **Comparison with stored answer** Stored correct answer: **C** Our derived answer: **C** So they agree.

More from Inverse Trigonometric Functions

View written solutionFree

Correct answer: C

Identify the pattern of the general term

The given series is

\cot^{-1}\left(\frac{7}{4}\right)+\cot^{-1}\left(\frac{19}{4}\right)+\cot^{-1}\left(\frac{39}{4}\right)+\cot^{-1}\left(\frac{67}{4}\right)+\cdots

Now observe the successive differences: $19-7=12,\quad 39-19=20,\quad 67-39=28$ Second difference is constant $8$ , so numerator is quadratic: $an^2+bn+c$ with $2a=8\Rightarrow a=4$ .

Let numerator be $4n^2+bn+c$ Using $n=1,2$ : $4+b+c=7 \Rightarrow b+c=3$ $16+2b+c=19 \Rightarrow 2b+c=3$ Subtracting gives $b=0$ , hence $c=3$ .

So the $n$ th term is $\cot^{-1}\left(\frac{4n^2+3}{4}\right).$

Thus $S=\sum_{n=1}^{\infty} \cot^{-1}\left(\frac{4n^2+3}{4}\right).$

Convert each term into a telescoping form

We use the identity $\cot^{-1}x-\cot^{-1}y=\cot^{-1}\left(\frac{xy+1}{y-x}\right)$ for suitable positive values.

Now compute $\cot^{-1}(2n-1)-\cot^{-1}(2n+1).$ Then

=\frac{4n^2-1+1}{2} =\frac{4n^2}{2}=2n^2.$$ That does not match. Instead try $$\cot^{-1}\left(\frac{2n-1}{2}\right)-\cot^{-1}\left(\frac{2n+1}{2}\right).$$ Using the same identity: $$x=\frac{2n-1}{2},\quad y=\frac{2n+1}{2}$$ Then $$xy+1=\frac{(2n-1)(2n+1)}{4}+1=\frac{4n^2-1}{4}+1=\frac{4n^2+3}{4}$$ and $$y-x=\frac{2n+1}{2}-\frac{2n-1}{2}=1.$$ Therefore $$\cot^{-1}\left(\frac{2n-1}{2}\right)-\cot^{-1}\left(\frac{2n+1}{2}\right) =\cot^{-1}\left(\frac{4n^2+3}{4}\right).$$ So each term becomes $$\cot^{-1}\left(\frac{4n^2+3}{4}\right)=\cot^{-1}\left(\frac{2n-1}{2}\right)-\cot^{-1}\left(\frac{2n+1}{2}\right).$$ --- 3. **Write the series as a telescoping sum** Hence

S=\sum_{n=1}^{\infty}\left[\cot^{-1}\left(\frac{2n-1}{2}\right)-\cot^{-1}\left(\frac{2n+1}{2}\right)\right].

Expanding,

S=\left[\cot^{-1}\left(\frac{1}{2}\right)-\cot^{-1}\left(\frac{3}{2}\right)\right] +\left[\cot^{-1}\left(\frac{3}{2}\right)-\cot^{-1}\left(\frac{5}{2}\right)\right] +\cdots

All middle terms cancel, so

S=\cot^{-1}\left(\frac{1}{2}\right)-\lim_{n\to\infty}\cot^{-1}\left(\frac{2n+1}{2}\right).

Since $$\lim_{n\to\infty}\cot^{-1}\left(\frac{2n+1}{2}\right)=0,$$ we get $$S=\cot^{-1}\left(\frac{1}{2}\right).$$ --- 4. **Match with the options** Use the identity for positive $x$: $$\cot^{-1}x=\frac{\pi}{2}-\tan^{-1}x.$$ Thus

\cot^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{2}-\tan^{-1}\left(\frac{1}{2}\right).

So the sum is $$\boxed{\frac{\pi}{2}-\tan^{-1}\left(\frac{1}{2}\right)}.$$ This is **Option C**. --- 5. **Comparison with stored answer** Stored correct answer: **C** Our derived answer: **C** So they agree.

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