Calculating Hubble's constant at earlier times I want to calculate Hubble's constant at some redshift $z$. I have found the following formula:
$$H^2=H_0^2\left(\Omega_m\left(1+z\right)^3+\Omega_{\Lambda}\right)$$
Now it's obvious that at higher redshift $z>0$ (earlier times) the Hubble's constant $H$ increases, which should not be the case. Why can't I use this formula and which formula should I use then?
 A: 
so, it's actually only numerically solvable? and is there no analytic expression for $H(z)$?

In my answer I'll be using the scale factor $a$ instead of the less wieldy redshift $z$. The two are simply related by $a = 1/(1+z)$.
In general the first Friedmann equation can be written as :
$$
H^2 = \left( \frac{\dot{a}}{a} \right)^2 = H_0^2 \left( \sum_i \Omega_i \times a^{-3(1+w_i)} + \Omega_k a^{-2} \right)
$$
Where $w$ is the equation of state parameter of each effective fluid $i$. Now let's take the case of a flat universe ($\Omega_k = 0$) where a single $\Omega_i$ dominates ($\Omega_j \approx 0$ for $j \neq i$). We get that :
\begin{eqnarray*}
\left( \frac{\dot{a}}{a} \right)^2 &\propto& a^{-3(1+w_i)} \\
\dot{a} &\propto& a^{-3/2(1+w)+1}
\end{eqnarray*}
Defining $\gamma$ such that $a \propto t^\gamma$ we get :
\begin{eqnarray*}
t^{\gamma - 1} &\propto& t^{-3/2 (1 + w) \gamma + \gamma} \\
\gamma - 1 &=& -3/2 (1 + w) \gamma + \gamma \\
\gamma &=& \frac{2}{3 (1+w)}
\end{eqnarray*}
This analytic solution is valid for all $w$ except when $w = -1$. There you get an exponential solution instead, and I'll leave it as an exercise for the reader to convince himself of it. As an example, for a flat matter dominated universe ($\Omega_m = 1$ and $\Omega_k = 0$) we get $a \propto t^{2/3}$ because $w_m = 0$.
What happens when we're not in such a simple scenario and there is more than one contributing effective fluid ? In that case there the analytic solutions are more intricate, and a guide to some of them may be found here
Note that you had an analytic form for $H(a)$ right from the start, and if you want one for $H(t)$ then it's pretty straightforward to plug in the solution I just gave you into the Friedmann equation.
