Σ_(n=1) ^∞ ((n!)/3^(n+1) )


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Question Number 116059 by Study last updated on 30/Sep/20

$\underset{n = 1}{\sum^{\infty}} \frac{n!}{3^{n + 1}}$
	Answered by MWSuSon last updated on 30/Sep/20

	$Do you want to test for convergence ?$
	Answered by mathmax by abdo last updated on 30/Sep/20
	$let s(x) =Σ_(n=1) ^∞ n! x^(n+1) ⇒s^′ (x) =Σ_(n=1) ^∞ (n+1)! x^n =Σ_(n=2) ^∞ n! x^(n−1) =(1/x^2 )Σ_(n=2) ^∞ n! x^(n+1) =(1/x^2 ){Σ_(n=1) ^∞ n! x^(n+1) −x^2 }=(1/x^2 )s(x)−1 ⇒ s^′ (x) =((s(x))/x^2 )−1 ⇒x^2 s^′ (x) =s(x)−x^2 ⇒x^2 s^′ (x)−s(x)+x^2 =0 (∣x∣<1) let solve x^2 y^′ −y =−x^2 h→x^2 y^′ =y ⇒(y^′ /y) =(1/x^2 ) ⇒ln∣y∣ =−(1/x) +c ⇒y =k e^(−(1/x)) lagrange method →y^′ =k^′ e^(−(1/x)) +k((1/x^2 ))e^(−(1/x)) e⇒x^2 k^′ e^(−(1/x)) +k e^(−(1/x)) −ke^(−(1/x)) =−x^2 ⇒k^′ e^(−(1/x)) =−1 ⇒ k^′ =−e^(1/x) ⇒k(x) =−∫_x ^1 e^(1/x) dx+c ⇒ s(x) =e^(−(1/x)) { c−∫_x ^1 e^(1/x) dx } ⇒s((1/3)) =e^(−(1/3)) {c−∫_(1/3) ^1 e^(1/x) dx} ....be continued...$
	$let s (x) = \sum_{n = 1}^{\infty} n! x^{n + 1} \Rightarrow s^{^{'}} (x) = \sum_{n = 1}^{\infty} (n + 1)! x^{n} = \sum_{n = 2}^{\infty} n! x^{n - 1}$ $= \frac{1}{x^{2}} \sum_{n = 2}^{\infty} n! x^{n + 1} = \frac{1}{x^{2}} {\sum_{n = 1}^{\infty} n! x^{n + 1} - x^{2}} = \frac{1}{x^{2}} s (x) - 1 \Rightarrow$ $s^{^{'}} (x) = \frac{s (x)}{x^{2}} - 1 \Rightarrow x^{2} s^{^{'}} (x) = s (x) - x^{2} \Rightarrow x^{2} s^{^{'}} (x) - s (x) + x^{2} = 0 (∣ x ∣< 1)$ $let solve x^{2} y^{^{'}} - y = - x^{2}$ $h \to x^{2} y^{^{'}} = y \Rightarrow \frac{y^{^{'}}}{y} = \frac{1}{x^{2}} \Rightarrow \ln ∣ y ∣ = - \frac{1}{x} + c \Rightarrow y = k e^{- \frac{1}{x}}$ $lagrange method \to y^{^{'}} = k^{^{'}} e^{- \frac{1}{x}} + k (\frac{1}{x^{2}}) e^{- \frac{1}{x}}$ $e \Rightarrow x^{2} k^{^{'}} e^{- \frac{1}{x}} + k e^{- \frac{1}{x}} - {ke}^{- \frac{1}{x}} = - x^{2} \Rightarrow k^{^{'}} e^{- \frac{1}{x}} = - 1 \Rightarrow$ $k^{^{'}} = - e^{\frac{1}{x}} \Rightarrow k (x) = - \int_{x}^{1} e^{\frac{1}{x}} dx + c \Rightarrow$ $s (x) = e^{- \frac{1}{x}} {c - \int_{x}^{1} e^{\frac{1}{x}} dx} \Rightarrow s (\frac{1}{3}) = e^{- \frac{1}{3}} {c - \int_{\frac{1}{3}}^{1} e^{\frac{1}{x}} dx}$ $. . . . be continued . . .$
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