Question

show that (1+1/2!+1/4!+...)^2=1+(1+1/3!+1/5!+...)^2
$(1+1/2!+1/4!+...) ^{2} =1+(1+1/3!+1/5!+...)^2$
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Answer 1

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1+3!1+5!1+7!1+...∞2!1+4!1+6!1+...∞

=2{1+3!1+5!1+...∞}2{2!1+4!1+6!1+...∞}=(e−e−1)(e+e−1)−2

=e−e1e+e1−2

=e2−1