...nice calculus... evaluate ::: lim_(x→0) {(1/x)[((ln(Γ(1+x))/x)−ψ(x+1)]}=?


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Question Number 124548 by mnjuly1970 last updated on 04/Dec/20
$...nice calculus... evaluate ::: lim_(x→0) {(1/x)[((ln(Γ(1+x))/x)−ψ(x+1)]}=?$
$. . . n i c e c a l c u l u s . . .$ $e v a l u a t e :::$ ${l i m}_{x \to 0} {\frac{1}{x} [\frac{l n (Γ (1 + x)}{x} - ψ (x + 1)]} = ?$
	Answered by mindispower last updated on 04/Dec/20
	$ln(Γ(1+x))=f(x) f(x)=f(0)+f′(0)x+((f′′(0)x^2 )/2)+o(x^2 ) f(0)=0,f′(0)=Ψ(1),f′′(0)=Ψ^1 (1) Ψ(x+1)=Ψ(1)+Ψ^1 (1)x+o(x) ((ln(Γ(1+x)))/x)−Ψ(x+1)=((xΨ(1)+(x^2 /2)Ψ^1 (1)+o(x^2 ))/x)−Ψ(1)−Ψ^1 (1)x+o(x) =−((Ψ^1 (1))/2)x+o(x) lim_(x→0) {(1/x)[((ln(Γ(1+x)))/x)−Ψ(1+x)]}=lim_(x→0) (1/x).((−Ψ(1))/2)x+o(x) =lim_(x→0) −((Ψ^1 (1))/2)+o(1)=−((Ψ^1 (1))/2) Ψ^1 (z)=Σ_(j≥0) (1/((j+z)^2 ))⇒Ψ^1 (1)=Σ_(j≥0) (1/((1+j)^2 ))=(π^2 /6) we get −(π^2 /(12))$
	$l n (Γ (1 + x)) = f (x)$ $f (x) = f (0) + f^{'} (0) x + \frac{f^{″} (0) x^{2}}{2} + o (x^{2})$ $f (0) = 0, f^{'} (0) = Ψ (1), f^{″} (0) = Ψ^{1} (1)$ $Ψ (x + 1) = Ψ (1) + Ψ^{1} (1) x + o (x)$ $\frac{l n (Γ (1 + x))}{x} - Ψ (x + 1) = \frac{x Ψ (1) + \frac{x^{2}}{2} Ψ^{1} (1) + o (x^{2})}{x} - Ψ (1) - Ψ^{1} (1) x + o (x)$ $= - \frac{Ψ^{1} (1)}{2} x + o (x)$ $\lim_{x \to 0} {\frac{1}{x} [\frac{l n (Γ (1 + x))}{x} - Ψ (1 + x)]} = \lim_{x \to 0} \frac{1}{x} . \frac{- Ψ (1)}{2} x + o (x)$ $= \lim_{x \to 0} - \frac{Ψ^{1} (1)}{2} + o (1) = - \frac{Ψ^{1} (1)}{2}$ $Ψ^{1} (z) = \sum_{j ⩾ 0} \frac{1}{{(j + z)}^{2}} \Rightarrow Ψ^{1} (1) = \sum_{j ⩾ 0} \frac{1}{{(1 + j)}^{2}} = \frac{π^{2}}{6}$ $w e g e t - \frac{π^{2}}{12}$
	Commented by mnjuly1970 last updated on 04/Dec/20

	$b r a v o m r m i n d s p o w e r$ $m a c l a u r e n e x p a n s i o n o f Γ (x + 1)$ $g r a t e f u l . . .$
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