jmc

$f^{-1}\left(\frac{1}{5}\right)$ is defined as the number $x$ such that $f(x)=\frac{1}{5}$. Thus, we solve the equation $$\frac{2}{x+1}=\frac{1}{5}.$$Multiplying both sides by $5(x+1)$, we have $$10=x+1.$$Subtracting $1$ from both sides gives $x=\boxed{9}$.