I am wondering, are these properties of modified Bessel functions correct?
$\frac{I_{1}(z)}{I_{o}(z)}\approx 1$
and
$\frac{I_{0}(2z)}{I_{o}^{2}(z)}\approx z^{1/2}\pi^{1/2},$
for large $z.$ If so, does somebody have an idea how to show that, or recommend me some good literature? I used expansions, but didn't get me where I wanted.
Using the asymptotic expansion given in Winther's link, i.e.
$$I_a(z) = \frac{e^z}{\sqrt{2\pi z}}\left(1 - \frac{4a^2-1}{8z}+ O(z^{-2})\right) $$
You get $$I_0(z) = \frac{e^z}{\sqrt{2\pi z}}\left(1 + \frac{1}{8z}+ O(z^{-2})\right) $$ $$I_1(z) = \frac{e^{z}}{\sqrt{2\pi z}}\left(1 - \frac{3}{8z}+ O(z^{-2})\right) $$ $$I_0(z)^2 = \frac{e^{2z}}{4\pi z}\left(1 + \frac{1}{8z}+ O(z^{-2})\right)^2 $$ $$I_0(2z) = \frac{e^{2z}}{\sqrt{4\pi z}}\left(1 + \frac{1}{16z}+ O(z^{-2})\right)$$ and therefore $$\frac{I_0(z)}{I_1(z)}= 1 + \frac{1}{2z} + O(z^{-2})$$ $$\frac{I_0(2z)}{I_0(z)^2}= \sqrt{\pi z} - \frac{3}{16}\sqrt{\frac{\pi}{z}} + O(z^{-3/2})$$ So your approximations are the correct first order asymptotic terms.
Edit: Here is a short derivation: Writing $x=1/z$ and omitting the exponential prefactors you get for large $z$, i.e. small $x$ using the geometric series $$\frac{1}{1-a} = 1 + a + a^2 + \cdots$$ $$\frac{I_0(z)}{I_1(z)} \approx \frac{1+\frac{x}{8}}{1-\frac{3x}{8}} \approx (1+\frac{x}{8})(1+\frac{3x}{8}) \approx 1 + \frac{x}{2}$$ $$\frac{I_0(2z)}{I_0(z)^2} \approx \frac{1+\frac{x}{16}}{(1+\frac{x}{8})^2} = \frac{1+\frac{x}{16}}{1+\frac{x}{4}+\frac{x^2}{64}} \approx (1+\frac{x}{16})(1-\frac{x}{4}) \ \approx 1 - \frac{3x}{16}$$
I have calculated the inverse of your first quotient, so $$\frac{I_1(z)}{I_0(z)}= 1 - \frac{1}{2z} + O(z^{-2})$$