Question Number 140978 by Mathspace last updated on 14/May/21
$${calculate}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left({n}!\right)^{\mathrm{2}} } \\ $$
Answered by Dwaipayan Shikari last updated on 14/May/21
$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!{n}!}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{1}\right)_{{n}} ^{\mathrm{2}} }{\left(\mathrm{1}\right)_{{n}} ^{\mathrm{3}} {n}!}\left(\mathrm{1}\right)^{{n}} =\:_{\mathrm{2}} {F}_{\mathrm{3}} \left(\mathrm{1},\mathrm{1};\mathrm{1};\mathrm{1};\mathrm{1};\mathrm{1}\right) \\ $$