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Question Number 206890 by mathzup last updated on 29/Apr/24

$$cansomeonefindtheexactvalueof$$$$\sum _{n=0}^{\mathrm{\infty }}\frac{1}{{(n!)}^2}$$

Commented by Frix last updated on 30/Apr/24

$$\mathrm{This}\mathrm{is}\mathrm{a}\mathrm{Modified}\mathrm{Bressel}\mathrm{Function}\mathrm{of}\mathrm{the}$$$$\mathrm{First}\mathrm{Kind}.$$$$I_n\left(z\right)=\frac{1}{\pi }\underset{0}{\stackrel{\pi }{\int }}{\mathrm{e}}^{z\cos \theta }\cos n\theta d\theta$$$$\mathrm{With}n=0\mathrm{this}\mathrm{is}\mathrm{equal}\mathrm{to}$$$$I_0\left(z\right)=\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{{\left(\frac{z^2}{4}\right)}^k}{{(k!)}^2}$$$$\Rightarrow \underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{(k!)}^2}=I_0\left(2\right)\approx 2.27958530234$$$$\mathrm{I}\mathrm{found}\mathrm{this}\mathrm{in}\mathrm{a}\mathrm{script},\mathrm{without}\mathrm{proof},\mathrm{so}$$$${\mathrm{you}}^{\prime }\mathrm{ll}\mathrm{have}\mathrm{to}\mathrm{do}\mathrm{some}\mathrm{research}...$$

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