6. Given the function $f(x)=\frac{(\sqrt{2})^{x}-1}{(\sqrt{2})^{x}+1}$, the smallest positive integer $n$ such that $f(n)>\frac{n}{n+1}$ is $\qquad$
6. 9. $f(n)>\frac{n}{n+1}$, that is, $\frac{(\sqrt{2})^{n}-1}{(\sqrt{2})^{n}+1}>\frac{n}{n+1}$, which also means $1-\frac{2}{(\sqrt{2})^{n}+1}>1-\frac{1}{n+1}, (\sqrt{2})^{n}>2n+1$.