.......advanced calculus...... prove that:: lim_(n→∞) {(((−1)^(n+1) n^(n+1) )/(n!)) (d^( n) /dx^n )(((ln(x))/x))∣


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Question Number 139530 by mnjuly1970 last updated on 28/Apr/21
$.......advanced calculus...... prove that:: lim_(n→∞) {(((−1)^(n+1) n^(n+1) )/(n!)) (d^( n) /dx^n )(((ln(x))/x))∣_(x=n) }=γ γ : euler −mascheroni constant$
$. . . . . . . a d v a n c e d c a l c u l u s . . . . . .$ $p r o v e t h a t ::$ ${l i m}_{n \to \infty} {\frac{{(- 1)}^{n + 1} n^{n + 1}}{n!} \frac{d^{n}}{{d x}^{n}} (\frac{l n (x)}{x}) ∣_{x = n}} = γ$ $γ : e u l e r - m a s c h e r o n i c o n s t a n t$
	Answered by mindispower last updated on 28/Apr/21

	$\frac{d^{n}}{{d x}^{n}} \frac{l n (x)}{x} = \underset{k = 0}{\sum^{n}} C_{n}^{k} {(l n (x))}^{k} . {(\frac{1}{x})}^{n - k} . L i b n e i z f o r m u l a$ $l n {(x)}^{(k)} = l n (x), k = 0$ $= {(\frac{1}{x})}^{(k - 1)}, k ⩾ 1$ $= \frac{{(- 1)}^{k - 1} . (k - 1)!}{x^{k - 1}}, k ⩾ 1$ ${(\frac{1}{x})}^{n - k} = \frac{{(- 1)}^{n - k} (n - k)!}{x^{n - k + 1}}$ $= (\underset{k = 1}{\sum^{n}} C_{n}^{k} \frac{{(- 1)}^{n - 1} (k - 1)! . (n - k)!}{x^{n + 1}} + l n (x) . \frac{{(- 1)}^{n} n!}{x^{n + 1}}) ∣_{x = n} . \frac{{(- 1)}^{n + 1} n^{n + 1}}{n!}$ $= \sum_{k ⩾ 1} \frac{{(- 1)}^{n - 1}}{n^{n + 1}} . \frac{(k - 1)! . (n - k)!}{n^{n + 1}} . \frac{n!}{k! . (n - k)!} . \frac{{(- 1)}^{n + 1}}{n!} n^{n + 1}$ $+ l n (n) . \frac{{(- 1)}^{n}}{n^{n + 1}} n! . \frac{{(- 1)}^{n + 1} n^{n + 1}}{n!}$ $= \underset{k = 1}{\sum^{n}} \frac{1}{k} - l n (n)$ $\lim_{n \to \infty} \underset{k = 1}{\sum^{n}} \frac{1}{k} - l n (n) = γ B y d e f i n i t i o n$
	Commented by mnjuly1970 last updated on 28/Apr/21

	$t h a n k s a l o t s i r p o w e r . . .$ $m e r c e y . . . .$
	Commented by mindispower last updated on 29/Apr/21

	$p l e a s u r$
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