Note that $$\ln(2g(t))=\frac{1}{2} \,\ln \left(1-t^2\right)+ t \tanh ^{-1}(t) =\sum_{k=1}^\infty\frac{t^{2k}}{2k(2k-1)}.$$ This immediately yields the positive answer to Question 2.
Iosif Pinelis
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