lim U_n =Σ_(k=o) ^(n−1) ((n(ln(n+k))−ln(n))/(n^2 +k^2 ))


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Question Number 155310 by SANOGO last updated on 28/Sep/21

$\lim U_{n} = \underset{k = o}{\sum^{n - 1}} \frac{n (l n (n + k)) - l n (n)}{n^{2} + k^{2}}$
	Answered by puissant last updated on 28/Sep/21

	$\lim_{x \to \infty} U_{n} = \underset{k = 0}{\sum^{n - 1}} \frac{n (l n (\frac{n + k}{n}))}{n^{2} + k^{2}}$ $\Rightarrow \lim_{x \to \infty} U_{n} = \lim_{x \to \infty} \frac{1}{n} \underset{k = 0}{\sum^{n - 1}} \frac{l n (1 + (\frac{k}{n}))}{1 + {(\frac{k}{n})}^{2}}$ $q u i e s t s o u s l a f o r m e \lim_{x \to \infty} \frac{b - a}{n} \underset{k = 0}{\sum^{n - 1}} f (a + k \frac{b - a}{n})$ $q u i e s t u n e I n t e g r a l e d e R i e m a n n, e t d o n n e a l o r s :$ $\lim_{x \to \infty} U_{n} = \int_{0}^{1} \frac{l n (1 + x)}{1 + x^{2}} d x = Q$ $x = t a n t \to Q = \int_{0}^{\frac{π}{4}} \frac{l n (1 + t a n t)}{(1 + {t a n}^{2} t)} (1 + {t a n}^{2} t) d t$ $= \int_{0}^{\frac{π}{4}} l n (1 + t a n t) d t; u = \frac{π}{4} - t \to d t = - d u$ $\Rightarrow Q = \int_{0}^{\frac{π}{4}} l n (\frac{2}{1 + t a n u}) d u = \int_{0}^{\frac{π}{4}} l n 2 d u - Q$ $\Rightarrow 2 Q = \int_{0}^{\frac{π}{4}} l n 2 d u \Rightarrow Q = \frac{π}{8} l n 2 . . .$ $\lim_{x \to \infty} U_{n} = \underset{k = 0}{\sum^{n - 1}} \frac{n (l n (n + k) - l n (n))}{n^{2} + k^{2}} = \frac{π}{8} l n 2 . .$ $∵∴ . . . . . . L e p u i s s a n t . . . . . . . .$
	Commented by SANOGO last updated on 28/Sep/21

	$t u e s t v r a i m e n t p u i s s a n t m e r c i$
	Commented by Tawa11 last updated on 28/Sep/21

	$nice sir$
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