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

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.

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

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