Question Number 206890 by mathzup last updated on 29/Apr/24 | ||
$${can}\:{some}\:{one}\:{find}\:{the}\:{exact}\:{value}\:{of} \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left({n}!\right)^{\mathrm{2}} } \\ $$ | ||
Commented by Frix last updated on 30/Apr/24 | ||
$$\mathrm{This}\:\mathrm{is}\:\mathrm{a}\:\mathrm{Modified}\:\mathrm{Bressel}\:\mathrm{Function}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{First}\:\mathrm{Kind}. \\ $$$${I}_{{n}} \:\left({z}\right)\:=\frac{\mathrm{1}}{\pi}\underset{\mathrm{0}} {\overset{\pi} {\int}}\mathrm{e}^{{z}\mathrm{cos}\:\theta} \mathrm{cos}\:{n}\theta\:{d}\theta \\ $$$$\mathrm{With}\:{n}=\mathrm{0}\:\mathrm{this}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$$${I}_{\mathrm{0}} \:\left({z}\right)\:=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{{z}^{\mathrm{2}} }{\mathrm{4}}\right)^{{k}} }{\left({k}!\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\:\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\left({k}!\right)^{\mathrm{2}} }\:={I}_{\mathrm{0}} \:\left(\mathrm{2}\right)\:\approx\mathrm{2}.\mathrm{27958530234} \\ $$$$\mathrm{I}\:\mathrm{found}\:\mathrm{this}\:\mathrm{in}\:\mathrm{a}\:\mathrm{script},\:\mathrm{without}\:\mathrm{proof},\:\mathrm{so} \\ $$$$\mathrm{you}'\mathrm{ll}\:\mathrm{have}\:\mathrm{to}\:\mathrm{do}\:\mathrm{some}\:\mathrm{research}... \\ $$ | ||