prove that lim_(n→∞) nΣ_(k=1) ^(n−1) ((ln(k+n)−ln(n))/(k^2 +n^2 ))=(π/8)ln2


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Question Number 111879 by mathdave last updated on 05/Sep/20

$p r o v e t h a t$ $\lim_{n \to \infty} n \underset{k = 1}{\sum^{n - 1}} \frac{\ln (k + n) - \ln (n)}{k^{2} + n^{2}} = \frac{π}{8} \ln 2$
	Answered by Dwaipayan Shikari last updated on 05/Sep/20

	$\lim_{n \to \infty} n \underset{k = 0}{\sum^{n}} \frac{l o g (\frac{k}{n} + 1)}{k^{2} + n^{2}} = \lim_{n \to \infty} \frac{1}{n} \underset{k = 0}{\sum^{n}} \frac{l o g (\frac{k}{n} + 1)}{\frac{k^{2}}{n^{2}} + 1}$ $= \int_{0}^{1} \frac{l o g (x + 1)}{x^{2} + 1} d x$ $= \int_{0}^{\frac{π}{4}} \frac{l o g (t a n θ + 1)}{{t a n}^{2} θ + 1} . {s e c}^{2} θ d θ$ $= \int_{0}^{\frac{π}{4}} l o g (s i n θ + c o s θ) - l o g (c o s θ)$ $= \int_{0}^{\frac{π}{4}} l o g (\sqrt{2}) + l o g (\frac{s i n θ + c o s θ}{\sqrt{2}}) - l o g (c o s θ) d θ$ $= \int_{0}^{\frac{π}{4}} \frac{1}{2} l o g (2) + [\int_{0}^{\frac{π}{4}} l o g (c o s (\frac{π}{4} - θ) - l o g (c o s θ))] = I$ $= \frac{π}{8} l o g (2) + I = \frac{π}{8} l o g (2)$ $((\int_{0}^{\frac{π}{4}} l o g (c o s (\frac{π}{4} - θ)) - l o g (c o s θ) = \int_{0}^{\frac{π}{4}} l o g (c o s θ) - l o g (c o s (\frac{π}{4} - θ)) = I$ $2 I = 0))$
	Commented by mathdave last updated on 05/Sep/20

	$o h n o c h e c k u r w o r k i n g a g a i n$
	Commented by mnjuly1970 last updated on 05/Sep/20

	$f a m o u s i n t e g r a l a n d i t s$ $a n s w e r : \frac{π}{8} l n (2) ✓$
	Commented by mathdave last updated on 05/Sep/20

	$t h i s u r w o r k i n g e n h h h j u s t d e y$ $s o m e h o w$
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