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Question Number 88569 by M±th+et£s last updated on 11/Apr/20

Commented by M±th+et£s last updated on 11/Apr/20

$p r o v e t h a t$ $∵ \emptyset l e r c h t r a n s c e n d e n t$
	Answered by mind is power last updated on 12/Apr/20
	$∅(x,1,a)=Σ_(k≥0) (x^k /((k+a))) ∫x^n ∅(x,1,a)dx=Σ_(k≥0) ∫(x^(k+n) /(k+a))dx =Σ_(k≥0) (x^(k+n+1) /((k+a)(k+n+1)))=Σx^(k+n+1) ((1/(n+1−a))((1/(k+a))−(1/(k+n+1)))) =(1/(a−1−n)){Σ(x^(k+n+1) /(k+n+1))−Σ(x^(k+n+1) /(k+a))} Σ(x^(k+n+1) /(k+n+1))=x^(n+1) ∅(x,1,n+1) Σ(x^(k+n+1) /(k+a))=x^(n+1) ∅(x,1,a) ((x^(n+1) ∅(x,1,n+1)−x^(n+1) ∅(x,1,a))/(1−a−n))$
	$\emptyset (x, 1, a) = \sum_{k ⩾ 0} \frac{x^{k}}{(k + a)}$ $\int x^{n} \emptyset (x, 1, a) d x = \sum_{k ⩾ 0} \int \frac{x^{k + n}}{k + a} d x$ $= \sum_{k ⩾ 0} \frac{x^{k + n + 1}}{(k + a) (k + n + 1)} = Σ x^{k + n + 1} (\frac{1}{n + 1 - a} (\frac{1}{k + a} - \frac{1}{k + n + 1}))$ $= \frac{1}{a - 1 - n} {Σ \frac{x^{k + n + 1}}{k + n + 1} - Σ \frac{x^{k + n + 1}}{k + a}}$ $Σ \frac{x^{k + n + 1}}{k + n + 1} = x^{n + 1} \emptyset (x, 1, n + 1)$ $Σ \frac{x^{k + n + 1}}{k + a} = x^{n + 1} \emptyset (x, 1, a)$ $\frac{x^{n + 1} \emptyset (x, 1, n + 1) - x^{n + 1} \emptyset (x, 1, a)}{1 - a - n}$
	Commented by M±th+et£s last updated on 12/Apr/20

	$g r e a t s o l u t i o n$
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