Assuming equal probability of the deck where the split is.
I can find the expected size of the smaller deck of card is $\frac{1+n}{2}$, but I am not sure how the find the expected ratio in a closed form of $n$.
Fundamentally, it would be something like $\frac{1}{n}\sum_{i=1}^n \frac{i}{2n+1-i}$. Not sure if there is a closed form. Maybe there is a different approach.
I doubt there's a fully closed form; but you can write it as $$ \frac1n \sum_{i=n+1}^{2n} \frac{2n+1-i}i = \frac{2n+1}n \sum_{i=n+1}^{2n} \frac1i - \frac1n \sum_{i=n+1}^{2n} 1 = \frac{2n+1}n (H_{2n}-H_n) - 1 $$ in terms of the harmonic numbers $H_m$. (As a side comment, note that the limit of the right-hand side as $n\to\infty$ equals $2\log 2-1$, consistent with the question lulu pointed to in a comment.)