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Question Number 130181 by Adel last updated on 23/Jan/21

	Answered by Olaf last updated on 23/Jan/21
	$lim_(x→3) ((Γ(x+1)−6)/(xx−33)) = lim_(x→3) ((Γ(x+1)−6)/(11(x−3))) = lim_(x→3) ((Γ′(x+1))/(11)) = lim_(x→3) ((Γ(x+1)ψ(x+1))/(11)) = ((Γ(4)ψ(4))/(11)) = ((3!(H_3 −γ))/(11)) = ((6(((11)/6)−γ))/(11)) = 1−(6/(11))γ (1) ψ(3) = H_2 −γ = (3/2)−γ (2) S = Σ_(n=1) ^∞ ((((1/2))^n )/n) (1/(1−x)) = Σ_(n=0) ^∞ x^n −ln(1−x) = Σ_(n=0) ^∞ (x^(n+1) /(n+1)) = Σ_(n=1) ^∞ (x^n /n) For x = (1/2), S = ln2 (3) With (1), (2) and (3), the result is : {(1−(6/(11))γ)^((3/2)−γ) }^(ln2) ≈ {(1−(6/(11))0,5772156649)^((3/2)−0,5772156649) }^(ln2) ≈ 0,7851749027$
	$\lim_{x \to 3} \frac{Γ (x + 1) - 6}{x x - 33}$ $= \lim_{x \to 3} \frac{Γ (x + 1) - 6}{11 (x - 3)}$ $= \lim_{x \to 3} \frac{Γ^{'} (x + 1)}{11}$ $= \lim_{x \to 3} \frac{Γ (x + 1) ψ (x + 1)}{11}$ $= \frac{Γ (4) ψ (4)}{11}$ $= \frac{3! (H_{3} - γ)}{11}$ $= \frac{6 (\frac{11}{6} - γ)}{11}$ $= 1 - \frac{6}{11} γ (1)$ $ψ (3) = H_{2} - γ = \frac{3}{2} - γ (2)$ $S = \underset{n = 1}{\sum^{\infty}} \frac{{(\frac{1}{2})}^{n}}{n}$ $\frac{1}{1 - x} = \underset{n = 0}{\sum^{\infty}} x^{n}$ $- \ln (1 - x) = \underset{n = 0}{\sum^{\infty}} \frac{x^{n + 1}}{n + 1} = \underset{n = 1}{\sum^{\infty}} \frac{x^{n}}{n}$ $For x = \frac{1}{2}, S = \ln 2 (3)$ $With (1), (2) and (3), the result is :$ ${{(1 - \frac{6}{11} γ)}^{\frac{3}{2} - γ}}^{\ln 2}$ $\approx {{(1 - \frac{6}{11} 0, 5772156649)}^{\frac{3}{2} - 0, 5772156649}}^{\ln 2}$ $\approx 0, 7851749027$
	Commented by Adel last updated on 23/Jan/21

	$thanks sir$
	Commented by 0731619177 last updated on 24/Jan/21

	$h h h h h h h$ څه وایی تره بچه
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