Question

\(\sum\limits_{n = 1}^\infty {\arctan (n)} \) Find Whether It Is Convergent Or Divergent And Find Its Sum If It Is Convergent.

Short answer

Given \(\) \(\sum\limits_{n = 1}^\infty {\arctan (n) = \sum\limits_{n = 1}^\infty {{{\tan }^{ - 1}}} } (n)\)

Step-by-step solution

Find Limit Of The \({n^{th}}\) Term

\( \Rightarrow \mathop {\lim }\limits_{n \to \infty } {\tan ^{ - 1}}(n) = \mathop {\lim }\limits_{n \to \infty } {\tan ^{ - 1}}(n)\)

\(\begin{aligned} &= {\tan ^{ - 1}}(\infty )\\ &= ta{n^{ - 1}}(tan\frac{\pi }{2})\end{aligned}\) \(\)

\( \Rightarrow \mathop {\lim }\limits_{n \to \infty } {\tan ^{ - 1}}(n) = \frac{\pi }{2}\)

Divergence Series Test

ByDivergence Series Test\(\mathop {\lim }\limits_{n \to \infty } {\tan ^{ - 1}}(n) = \frac{\pi }{2} \ne 0\)

Hence, the \({\tan ^{ - 1}}(n)\) Diverges and it is a Divergent Series.