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Question Number 72746 by aliesam last updated on 01/Nov/19
Σ_(n=0) ^∞ (1/((5n)!))
$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{5}{n}\right)!} \\ $$
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)
$$\mathrm{1}+\mathrm{z}+\mathrm{z}^{\mathrm{2}} +\mathrm{z}^{\mathrm{3}} +\mathrm{z}^{\mathrm{4}} =\mathrm{0}\Rightarrow\mathrm{Z}=\mathrm{e}^{\frac{\mathrm{2ik}\pi}{\mathrm{5}}} ,\mathrm{k}\in\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4}\right\}\: \\ $$$$\mathrm{e}^{\mathrm{e}^{\mathrm{i}\frac{\mathrm{2}\pi}{\mathrm{5}}} } =\Sigma\frac{\mathrm{e}^{\frac{\mathrm{2ik}\pi}{\mathrm{5}}} }{\mathrm{k}!}=\underset{\mathrm{r}=\mathrm{0}} {\overset{\mathrm{4}} {\sum}}\underset{\mathrm{k}=\mathrm{0}} {\overset{+\infty} {\sum}}\frac{\mathrm{e}^{\frac{\mathrm{2i}\pi}{\mathrm{5}}\mathrm{r}} }{\left(\mathrm{5k}+\mathrm{r}\right)!} \\ $$$$\mathrm{e}^{\mathrm{e}^{\frac{\mathrm{i4}\pi}{\mathrm{5}}} } =\underset{\mathrm{k}\geqslant\mathrm{0}} {\overset{} {\sum}}\frac{\mathrm{e}^{\frac{\mathrm{4ik}\pi}{\mathrm{5}}} }{\mathrm{k}!}=\underset{\mathrm{r}=\mathrm{0}} {\overset{\mathrm{4}} {\sum}}\underset{\mathrm{k}=\mathrm{0}} {\overset{+\infty} {\sum}}\frac{\mathrm{e}^{\frac{\mathrm{4i}\pi\mathrm{r}}{\mathrm{5}}} }{\left(\mathrm{5k}+\mathrm{r}\right)!} \\ $$$$\mathrm{e}^{\mathrm{e}^{\frac{\mathrm{6i}\pi}{\mathrm{5}}} } =\Sigma\frac{\mathrm{e}^{\frac{\mathrm{6ik}\pi}{\mathrm{5}}} }{\mathrm{k}!}=\underset{\mathrm{r}=\mathrm{0}} {\overset{\mathrm{4}} {\sum}}\underset{\mathrm{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{e}^{\frac{\mathrm{6ir}\pi}{\mathrm{5}}} }{\left(\mathrm{5k}+\mathrm{r}\right)!} \\ $$$$\mathrm{e}^{\mathrm{e}^{\frac{\mathrm{8i}\pi}{\mathrm{5}}} } =\Sigma\frac{\mathrm{e}^{\frac{\mathrm{8ik}\pi}{\mathrm{5}}} }{\mathrm{k}!}=\underset{\mathrm{r}=\mathrm{0}} {\overset{\mathrm{4}} {\sum}}\underset{\mathrm{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{e}^{\frac{\mathrm{8ir}\pi}{\mathrm{5}}} }{\left(\mathrm{5k}+\mathrm{r}\right)!} \\ $$$$\mathrm{e}^{\mathrm{1}} =\underset{\mathrm{k}\geqslant\mathrm{0}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{k}!}=\underset{\mathrm{r}=\mathrm{0}} {\overset{\mathrm{4}} {\sum}}\underset{\mathrm{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{5k}+\mathrm{r}\right)!} \\ $$$$\Rightarrow\underset{\mathrm{j}=\mathrm{0}} {\overset{\mathrm{4}} {\sum}}\mathrm{e}^{\mathrm{e}^{\frac{\mathrm{2i}\pi\mathrm{j}}{\mathrm{5}}} } \:=\underset{\mathrm{r}=\mathrm{0}} {\overset{\mathrm{4}} {\sum}}\underset{\mathrm{k}\geqslant\mathrm{0}} {\sum}\frac{\underset{\mathrm{j}=\mathrm{0}} {\overset{\mathrm{4}} {\sum}}\mathrm{e}^{\frac{\mathrm{2ij}\pi\mathrm{r}}{\mathrm{5}}} }{\left(\mathrm{5k}+\mathrm{r}\right)!} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{if}\:\mathrm{r}\neq\mathrm{0} \\ $$$$\underset{\mathrm{j}=\mathrm{0}} {\overset{\mathrm{4}} {\sum}}\mathrm{e}^{\frac{\mathrm{2ij}\pi\mathrm{r}}{\mathrm{5}}} =\mathrm{0}\:\:\mathrm{cause}\:=\frac{\mathrm{1}−\left(\mathrm{e}^{\mathrm{2i}\pi\mathrm{r}} \right)}{\mathrm{1}+\mathrm{e}^{\frac{\mathrm{i2}\pi\mathrm{r}}{\mathrm{5}}} } \\ $$$$\Rightarrow\underset{\mathrm{j}=\mathrm{0}} {\overset{\mathrm{4}} {\sum}}\mathrm{e}^{\mathrm{e}^{\frac{\mathrm{2i}\pi\mathrm{j}}{\mathrm{5}}} } \:=\underset{\mathrm{r}=\mathrm{0}} {\overset{\mathrm{4}} {\sum}}\underset{\mathrm{k}\geqslant\mathrm{0}} {\sum}\frac{\underset{\mathrm{j}=\mathrm{0}} {\overset{\mathrm{4}} {\sum}}\mathrm{e}^{\frac{\mathrm{2ij}\pi\mathrm{r}}{\mathrm{5}}} }{\left(\mathrm{5k}+\mathrm{r}\right)!}=\underset{\mathrm{r}=\mathrm{0}} {\overset{\mathrm{0}} {\sum}}\underset{\mathrm{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{5}}{\left(\mathrm{5k}+\mathrm{r}\right)!}=\mathrm{5}\underset{\mathrm{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{5k}\right)!} \\ $$$$\Rightarrow\underset{\mathrm{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left(\mathrm{5k}\right)!}=\frac{\mathrm{e}^{\mathrm{e}^{\frac{\mathrm{2i}\pi}{\mathrm{5}}} } +\mathrm{e}^{\mathrm{e}^{\frac{\mathrm{4i}\pi}{\mathrm{5}}} } +\mathrm{e}^{\mathrm{e}^{\frac{\mathrm{6i}\pi}{\mathrm{5}}} } +\mathrm{e}^{\mathrm{e}^{\frac{\mathrm{8i}\pi}{\mathrm{5}}} } +\mathrm{e}^{\mathrm{1}} }{\mathrm{5}} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{generslise}\:\mathrm{this} \\ $$$$\Rightarrow\underset{\mathrm{k}=\mathrm{0}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{ak}\right)!}=\frac{\underset{\mathrm{j}=\mathrm{0}} {\overset{\mathrm{a}−\mathrm{1}} {\sum}}\mathrm{e}^{\mathrm{e}^{\frac{\mathrm{2i}\pi\mathrm{j}}{\mathrm{a}}} } }{\mathrm{a}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by aliesam last updated on 01/Nov/19
god bless you sir
$${god}\:{bless}\:{you}\:{sir} \\ $$

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