If $x > 0$, let $r = 1/|1 + 2 i/x| = 1/\sqrt{1 + 4/x^2}$ and the $n$'th term on the right of the second sum is dominated by $2/((2n-1)r^{2n-1})$.
On the other hand, for large $n$ the $n$'th term in the first sum is approximately $\sqrt{\pi/(4n)} (x/(1+x^2)) (x^2/(1+x^2))^n$.
Since $$ \frac{x^2}{1+x^2} > \frac{1}{1+4/x^2} = \frac{x^2}{4 + x^2}$$
for all $x > 0$ we find that the $n$'th term in the second sum will go to $0$ asymptotically faster.
Caution: this does not say that for a particular tolerance $\epsilon > 0$ the partial sums achieve tolerance within $\epsilon$ faster.