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Question Number 66800 by mathmax by abdo last updated on 19/Aug/19

c a l c u l a t e U_{n} = \int_{\frac{1}{n}}^{n} \frac{a r c t a n (x)}{1 + x^{2}} d x

a n d d e t e r m i n e {l i m}_{n \to + \infty} U_{n}

2) f i n d n a t u r e o f Σ U_{n}

Commented by mathmax by abdo last updated on 21/Aug/19

1) b y p a r t s u^{^{'}} = \frac{1}{1 + x^{2}} a n d v = a r c t a n x \Rightarrow

U_{n} = {[{a r c t a n}^{2} x]}_{\frac{1}{n}}^{n} - \int_{\frac{1}{n}}^{n} a r c t a n (x) \frac{d x}{1 + x^{2}} = {a r c t a n}^{2} (n) - {a r c t a n}^{2} (\frac{1}{n})

- U_{n} \Rightarrow 2 U_{n} = {a r c t a n}^{2} (n) - {a r c t a n}^{2} (\frac{1}{n}) \Rightarrow

U_{n} = \frac{1}{2} {{a r c t a n}^{2} (n) - {a r c t a n}^{2} (\frac{1}{n})} \Rightarrow {l i m}_{n \to + \infty} U_{n} = \frac{1}{2} {(\frac{π}{2})}^{2}

= \frac{π^{2}}{8}

2) w e h a v e {l i m}_{n \to + \infty} U_{n} \neq 0 \Rightarrow Σ U_{n} d i v e r g e s .