Does this continued fraction, $C_f(n)$ has a closed form?
$$C_f(n)={1\over4n+{1\over12n+{1\over 20n+{1\over 28n+\cdots}}}}\tag1$$
$n\ge1.$
Yes?
So I did some mathematical experimental for a while and sort came up with this as a closed form
$${e+1\over e-1}-{2\over e-1}\left(\sum_{j=1}^{2n-1}(-1)^{j+1}\sqrt[2n]{e^j}\right)\tag2$$
but how can we show that $(2)$ is correct?