Question

The value of \[ \lim_{n\to \infty}\sum_{r=1}^{n}\cot^{-1}\left(r^2+\frac{3}{4}\right) \] is

• A. \(\cot^{-1}2\)
• B. \(\cot^{-1}\frac{1}{3}\)
• C. \(\tan^{-1}2\)
• D. \(\tan^{-1}\frac{1}{3}\)

Hint:

For sums involving inverse trigonometric functions, try to write each term as a difference of two consecutive inverse tangent terms so that the series becomes telescopic.

Answer 1

The Correct Option is C

Step 1: Convert the term into tangent inverse form.
We know that \[ \cot^{-1}x=\tan^{-1}\left(\frac{1}{x}\right) \] So, \[ \cot^{-1}\left(r^2+\frac{3}{4}\right) = \tan^{-1}\left(\frac{1}{r^2+\frac{3}{4}}\right) \] \[ = \tan^{-1}\left(\frac{4}{4r^2+3}\right) \]

Step 2: Express the term as a telescopic difference.
Observe that \[ \tan^{-1}\left(\frac{1}{r-\frac{1}{2}}\right) - \tan^{-1}\left(\frac{1}{r+\frac{1}{2}}\right) \] Using \[ \tan^{-1}a-\tan^{-1}b=\tan^{-1}\left(\frac{a-b}{1+ab}\right) \] we get \[ \frac{\frac{1}{r-\frac{1}{2}}-\frac{1}{r+\frac{1}{2}}}{1+\frac{1}{\left(r-\frac{1}{2}\right)\left(r+\frac{1}{2}\right)}} \] \[ = \frac{\frac{1}{r^2-\frac{1}{4}}}{1+\frac{1}{r^2-\frac{1}{4}}} \] \[ = \frac{1}{r^2+\frac{3}{4}} \] Hence, \[ \cot^{-1}\left(r^2+\frac{3}{4}\right) = \tan^{-1}\left(\frac{1}{r-\frac{1}{2}}\right) - \tan^{-1}\left(\frac{1}{r+\frac{1}{2}}\right) \]

Step 3: Apply telescoping sum.
Therefore, \[ \sum_{r=1}^{n}\cot^{-1}\left(r^2+\frac{3}{4}\right) \] \[ = \sum_{r=1}^{n} \left[ \tan^{-1}\left(\frac{1}{r-\frac{1}{2}}\right) - \tan^{-1}\left(\frac{1}{r+\frac{1}{2}}\right) \right] \] The terms cancel successively, giving \[ = \tan^{-1}2-\tan^{-1}\left(\frac{1}{n+\frac{1}{2}}\right) \]

Step 4: Take the limit.
As \[ n\to \infty, \] we have \[ \frac{1}{n+\frac{1}{2}}\to 0 \] So, \[ \tan^{-1}\left(\frac{1}{n+\frac{1}{2}}\right)\to 0 \] Hence, \[ \lim_{n\to \infty}\sum_{r=1}^{n}\cot^{-1}\left(r^2+\frac{3}{4}\right) = \tan^{-1}2 \]

Step 5: Final conclusion.
Therefore, \[ \boxed{\tan^{-1}2} \]