Σ_(n=0) ^∞ (1/((5n)!))


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Question Number 72746 by aliesam last updated on 01/Nov/19

$\underset{n = 0}{\sum^{\infty}} \frac{1}{(5 n)!}$
	Answered by mind is power last updated on 01/Nov/19
	$1+z+z^2 +z^3 +z^4 =0⇒Z=e^((2ikπ)/5) ,k∈{1,2,3,4} e^e^(i((2π)/5)) =Σ(e^((2ikπ)/5) /(k!))=Σ_(r=0) ^4 Σ_(k=0) ^(+∞) (e^(((2iπ)/5)r) /((5k+r)!)) e^e^((i4π)/5) =Σ_(k≥0) ^ (e^((4ikπ)/5) /(k!))=Σ_(r=0) ^4 Σ_(k=0) ^(+∞) (e^((4iπr)/5) /((5k+r)!)) e^e^((6iπ)/5) =Σ(e^((6ikπ)/5) /(k!))=Σ_(r=0) ^4 Σ_(k≥0) (e^((6irπ)/5) /((5k+r)!)) e^e^((8iπ)/5) =Σ(e^((8ikπ)/5) /(k!))=Σ_(r=0) ^4 Σ_(k≥0) (e^((8irπ)/5) /((5k+r)!)) e^1 =Σ_(k≥0) ^(+∞) (1/(k!))=Σ_(r=0) ^4 Σ_(k≥0) (1/((5k+r)!)) ⇒Σ_(j=0) ^4 e^e^((2iπj)/5) =Σ_(r=0) ^4 Σ_(k≥0) ((Σ_(j=0) ^4 e^((2ijπr)/5) )/((5k+r)!)) we have if r≠0 Σ_(j=0) ^4 e^((2ijπr)/5) =0 cause =((1−(e^(2iπr) ))/(1+e^((i2πr)/5) )) ⇒Σ_(j=0) ^4 e^e^((2iπj)/5) =Σ_(r=0) ^4 Σ_(k≥0) ((Σ_(j=0) ^4 e^((2ijπr)/5) )/((5k+r)!))=Σ_(r=0) ^0 Σ_(k≥0) (5/((5k+r)!))=5Σ_(k≥0) (1/((5k)!)) ⇒Σ_(k≥0) (1/((5k)!))=((e^e^((2iπ)/5) +e^e^((4iπ)/5) +e^e^((6iπ)/5) +e^e^((8iπ)/5) +e^1 )/5) we can generslise this ⇒Σ_(k=0) ^(+∞) (1/((ak)!))=((Σ_(j=0) ^(a−1) e^e^((2iπj)/a) )/a)$
	$1 + z + z^{2} + z^{3} + z^{4} = 0 \Rightarrow Z = e^{\frac{2 ik π}{5}}, k \in {1, 2, 3, 4}$ $e^{e^{i \frac{2 π}{5}}} = Σ \frac{e^{\frac{2 ik π}{5}}}{k!} = \underset{r = 0}{\sum^{4}} \underset{k = 0}{\sum^{+ \infty}} \frac{e^{\frac{2 i π}{5} r}}{(5 k + r)!}$ $e^{e^{\frac{i 4 π}{5}}} = \underset{k ⩾ 0}{\sum^{}} \frac{e^{\frac{4 ik π}{5}}}{k!} = \underset{r = 0}{\sum^{4}} \underset{k = 0}{\sum^{+ \infty}} \frac{e^{\frac{4 i π r}{5}}}{(5 k + r)!}$ $e^{e^{\frac{6 i π}{5}}} = Σ \frac{e^{\frac{6 ik π}{5}}}{k!} = \underset{r = 0}{\sum^{4}} \sum_{k ⩾ 0} \frac{e^{\frac{6 ir π}{5}}}{(5 k + r)!}$ $e^{e^{\frac{8 i π}{5}}} = Σ \frac{e^{\frac{8 ik π}{5}}}{k!} = \underset{r = 0}{\sum^{4}} \sum_{k ⩾ 0} \frac{e^{\frac{8 ir π}{5}}}{(5 k + r)!}$ $e^{1} = \underset{k ⩾ 0}{\sum^{+ \infty}} \frac{1}{k!} = \underset{r = 0}{\sum^{4}} \sum_{k ⩾ 0} \frac{1}{(5 k + r)!}$ $\Rightarrow \underset{j = 0}{\sum^{4}} e^{e^{\frac{2 i π j}{5}}} = \underset{r = 0}{\sum^{4}} \sum_{k ⩾ 0} \frac{\underset{j = 0}{\sum^{4}} e^{\frac{2 ij π r}{5}}}{(5 k + r)!}$ $we have if r \neq 0$ $\underset{j = 0}{\sum^{4}} e^{\frac{2 ij π r}{5}} = 0 cause = \frac{1 - (e^{2 i π r})}{1 + e^{\frac{i 2 π r}{5}}}$ $\Rightarrow \underset{j = 0}{\sum^{4}} e^{e^{\frac{2 i π j}{5}}} = \underset{r = 0}{\sum^{4}} \sum_{k ⩾ 0} \frac{\underset{j = 0}{\sum^{4}} e^{\frac{2 ij π r}{5}}}{(5 k + r)!} = \underset{r = 0}{\sum^{0}} \sum_{k ⩾ 0} \frac{5}{(5 k + r)!} = 5 \sum_{k ⩾ 0} \frac{1}{(5 k)!}$ $\Rightarrow \sum_{k ⩾ 0} \frac{1}{(5 k)!} = \frac{e^{e^{\frac{2 i π}{5}}} + e^{e^{\frac{4 i π}{5}}} + e^{e^{\frac{6 i π}{5}}} + e^{e^{\frac{8 i π}{5}}} + e^{1}}{5}$ $we can generslise this$ $\Rightarrow \underset{k = 0}{\sum^{+ \infty}} \frac{1}{(ak)!} = \frac{\underset{j = 0}{\sum^{a - 1}} e^{e^{\frac{2 i π j}{a}}}}{a}$
	Commented by aliesam last updated on 01/Nov/19

	$g o d b l e s s y o u s i r$
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