Θ:=(Σ_(n=1) ^∞ (n^4 /(2^n . n!)))^(1/2) =?


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Question Number 141811 by mnjuly1970 last updated on 23/May/21

$Θ := {(\underset{n = 1}{\sum^{\infty}} \frac{n^{4}}{2^{n} . n!})}^{\frac{1}{2}} = ?$
	Answered by qaz last updated on 23/May/21

	${(x D)}^{4}$ $= x D x D x D x D$ $= x D x D x (D + {x D}^{2})$ $= x D x [(D + {x D}^{2}) + x (D^{2} + D^{2} + {x D}^{3}]$ $= x D x (D + 3 {x D}^{2} + x^{2} D^{3})$ $= x [(D + 3 {x D}^{2} + x^{2} D^{3}) + x (D^{2} + 3 D^{2} + 3 {x D}^{3} + 2 {x D}^{3} + x^{2} D^{4})]$ $= x D + 7 x^{2} D^{2} + 6 x^{3} D^{3} + x^{4} D^{4}$ $Θ = {(\underset{n = 1}{\sum^{\infty}} \frac{n^{4}}{2^{n} n!})}^{2} = {({(x D)}^{4} (\underset{n = 0}{\sum^{\infty}} \frac{x^{n}}{n!} - 1))}^{1 / 2} ∣_{x = 1 / 2}$ $= {[{(x D)}^{4} (e^{x} - 1)]}^{1 / 2} ∣_{x = 1 / 2}$ $= {[(x + 7 x^{2} + 6 x^{3} + x^{4}) e^{x}]}^{1 / 2} ∣_{x = 1 / 2}$ $= {(\frac{1}{2} + \frac{7}{4} + \frac{6}{8} + \frac{1}{16})}^{1 / 2} \sqrt[4]{e}$ $= \frac{7}{4} \sqrt[4]{e}$
	Commented by mnjuly1970 last updated on 23/May/21

	$t h a n k y o u s o m u c h m r q a z . .$
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