Find Σ_(n=1) ^∞ (1/((3n)!))=?


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Question Number 99300 by mr W last updated on 20/Jun/20

$Find \underset{n = 1}{\sum^{\infty}} \frac{1}{(3 n)!} = ?$
	Answered by maths mind last updated on 20/Jun/20
	$Z^3 −1=0⇒z∈{1,e^((2iπ)/3) ,e^((4iπ)/3) } e^z =Σ_(n≥0) (z^n /(n!)),a=e^((2iπ)/3) ,b=e^((4iπ)/3) ,c=1 e^a +e^b +e^c =Σ_(n≥0) ((1+e^((2inπ)/3) +e^((4inπ)/3) )/(n!)) =Σ_(j=0) ^2 .Σ_(n≥0) ((1+e^((2i(3n+j)π)/3) +e^((4i(3n+j)π)/3) )/((3n+j)!)) =Σ_(n≥0) (3/((3n)!))+Σ_(n≥0) ((1+e^((2iπ)/3) +e^((4iπ)/3) )/((3n+1)!))+Σ((1+e^((4iπ)/3) +e^((8iπ)/3) )/((3n+2)!)) 1+e^((2iπ)/3) +e^((4iπ)/3) =0,((8π)/3)=2π+((2π)/3)⇒1+e^((4iπ)/3) +e^(((8π)/3)i) =0 ⇒e+e^e^((2iπ)/3) +e^e^((4iπ)/3) =3Σ_(n≥0) (1/((3n)!))=3(Σ_(n≥1) (1/((3n)!))+1) Σ_(n≥1) (1/((3n)!))=((e+e^e^((2iπ)/3) +e^e^(4((iπ)/3)) )/3)−1$
	$Z^{3} - 1 = 0 \Rightarrow z \in {1, e^{\frac{2 i π}{3}}, e^{\frac{4 i π}{3}}}$ $e^{z} = \sum_{n ⩾ 0} \frac{z^{n}}{n!}, a = e^{\frac{2 i π}{3}}, b = e^{\frac{4 i π}{3}}, c = 1$ $e^{a} + e^{b} + e^{c} = \sum_{n ⩾ 0} \frac{1 + e^{\frac{2 i n π}{3}} + e^{\frac{4 i n π}{3}}}{n!}$ $= \underset{j = 0}{\sum^{2}} . \sum_{n ⩾ 0} \frac{1 + e^{\frac{2 i (3 n + j) π}{3}} + e^{\frac{4 i (3 n + j) π}{3}}}{(3 n + j)!}$ $= \sum_{n ⩾ 0} \frac{3}{(3 n)!} + \sum_{n ⩾ 0} \frac{1 + e^{\frac{2 i π}{3}} + e^{\frac{4 i π}{3}}}{(3 n + 1)!} + Σ \frac{1 + e^{\frac{4 i π}{3}} + e^{\frac{8 i π}{3}}}{(3 n + 2)!}$ $1 + e^{\frac{2 i π}{3}} + e^{\frac{4 i π}{3}} = 0, \frac{8 π}{3} = 2 π + \frac{2 π}{3} \Rightarrow 1 + e^{\frac{4 i π}{3}} + e^{\frac{8 π}{3} i} = 0$ $\Rightarrow e + e^{e^{\frac{2 i π}{3}}} + e^{e^{\frac{4 i π}{3}}} = 3 \sum_{n ⩾ 0} \frac{1}{(3 n)!} = 3 (\sum_{n ⩾ 1} \frac{1}{(3 n)!} + 1)$ $\sum_{n ⩾ 1} \frac{1}{(3 n)!} = \frac{e + e^{e^{\frac{2 i π}{3}}} + e^{e^{4 \frac{i π}{3}}}}{3} - 1$
	Commented by mr W last updated on 20/Jun/20

	$t h a n k s s i r! l e t m e s t u d y!$
	Commented by maths mind last updated on 20/Jun/20

	$n i c e d a y s i r w i t h e p l e a s u r$
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