JEE MATHEMATICS MCQ: LIMITS

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Analyzing +4 | -1

The value of

\lim\limits_{n \to \infty } \,6\tan \left\{ {\sum\limits_{r = 1}^n {{{\tan }^{ - 1}}} \left( {\frac{1}{{{r^2} + 3r + 3}}} \right)} \right\}

is equal to:

View answer and explanation

Correct option: C

Option C (3) is correct because the limit is evaluated using the arctangent subtraction formula and telescoping series.

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