calculate lim_(n→+∞) n^3 Σ_(k=1) ^n (1/(n^4 +k^2 n^2 +k^4 )) .


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Question Number 34217 by abdo imad last updated on 02/May/18

$c a l c u l a t e {l i m}_{n \to + \infty} n^{3} \sum_{k = 1}^{n} \frac{1}{n^{4} + k^{2} n^{2} + k^{4}} .$
Commented by math khazana by abdo last updated on 04/May/18

$l e t p u t S_{n} = n^{3} \sum_{k = 1}^{n} \frac{1}{n^{4} + k^{2} n^{2} + k^{4}}$ $S_{n} = \frac{1}{n} \sum_{k = 1}^{n} \frac{1}{1 + \frac{k^{2}}{n^{2}} + \frac{k^{4}}{n^{4}}} s o S_{n} i s a R i e m a n s u m$ $a n d {l i m}_{n \to + \infty} S_{n} = \int_{0}^{1} \frac{d x}{1 + x^{2} + x^{4}}$ $x^{4} + x^{2} + 1 = {(x^{2} + 1)}^{2} - x^{2} = (x^{2} + 1 - x) (x^{2} + 1 + x)$ $l e t d e c o m p o s e F (x) = \frac{1}{(x^{2} + x + 1) (x^{2} - x + 1)}$ $F (x) = \frac{a x + b}{x^{2} + x + 1} + \frac{c x + d}{x^{2} - x + 1}$ $F (- x) = F (x) \Rightarrow \frac{- a x + b}{x^{2} - x + 1} + \frac{- c x + d}{x^{2} + x + 1} = F (x) \Rightarrow$ $c = - a a n d b = d \Rightarrow$ $F (x) = \frac{a x + b}{x^{2} + x + 1} + \frac{- a x + b}{x^{2} - x + 1}$ $F (0) = 1 = 2 b \Rightarrow b = \frac{1}{2} \Rightarrow$ $F (x) = \frac{1}{2} (\frac{2 a x + 1}{x^{2} + x + 1} + \frac{- 2 a x + 1}{x^{2} - x + 1})$ $F (1) = \frac{1}{3} = \frac{1}{2} (\frac{2 a + 1}{3} + \frac{- 2 a + 1}{1}) \Rightarrow$ $1 = \frac{1}{2} (2 a + 1 - 6 a + 3) = \frac{1}{2} (- 4 a + 4) = - 2 a + 2 \Rightarrow$ $- 2 a = - 1 \Rightarrow a = \frac{1}{2} \Rightarrow$ $F (x) = \frac{1}{2} (\frac{x + 1}{x^{2} + x + 1} - \frac{x - 1}{x^{2} - x + 1})$ $\int_{0}^{1} \frac{d x}{x^{4} + x^{2} + 1} = \frac{1}{4} \int_{0}^{1} \frac{2 x + 2}{x^{2} + x + 1} d x - \frac{1}{4} \int_{0}^{1} \frac{2 x - 2}{x^{2} + x + 1} d x$ $\int_{0}^{1} \frac{2 x + 2}{x^{2} + x + 1} d x = \int_{0}^{1} \frac{2 x + 1}{x^{2} + x + 1} d x + \int_{0}^{1} \frac{d x}{x^{2} + x + 1}$ $= {[l n (x^{2} + x + 1)]}_{0}^{1} + \int_{0}^{1} \frac{d x}{{(x + \frac{1}{2})}^{2} + \frac{3}{4}} (x + \frac{1}{2} = \frac{\sqrt{3}}{2} u$ $= l n (3) + \int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} \frac{1}{\frac{3}{4} (1 + u^{2})} \frac{\sqrt{3}}{2} d u$ $= l n (3) + \frac{4}{3} \frac{\sqrt{3}}{2} \int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} \frac{d u}{1 + u^{2}}$ $= l n (3) + \frac{2 \sqrt{3}}{3} (a r c t a n (\sqrt{3}) - a r c t a n (\frac{1}{\sqrt{3}})) . . . .$
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