RK4 Order of Global Truncation Error Confusion I'm currently trying to integrate an equation using an RK4 integrator and I'm getting confused with regards to the global truncation error. 
The problem consists of an integration from $t = 0$ seconds to $t = 258492$ seconds using a step-size of $2$ seconds. The question is: what would be the order of the error at the end of the integration? 
I know that the global truncation error of the RK4 integration scheme is $O(h^4)$, but I'm confused as to what step-size $h$ I should use here to calculate the order of the error.
 A: Let's first get our definitions straight, then the confusion should be almost gone.
Given an Interval $I=[t_0, t_0+T]$ and the following separation 
$$t_0<t_1<…<t_N=t_0+T$$ 
the step-size is defined as $h_n:= t_n-t_{n-1}$.
A one-step method 
$$u_n = u_{n-1}+h_nF(t_n,h_n;u_{n-1},u_n)$$
is called consistent of order $p$ if for the truncation error 
$$τ_n=h_n^{-1}(u(t_n)-u(t_{n-1}))-F(t_{n-1},h_n;u(t_{n-1}),u(t_n))$$
it holds
$$\max_{t_n∈I}\|τ_n\|=\mathcal{O}(h^p),$$
for $h=\max_{1≤n≤N}h_n→0$.
If the method applied to an ODE is also stable, then it converges with order $p$. The applied method is stable, if the right hand side $f$ of the ODE $u'=f$ fulfils some Lipschitz-condition. 
Keep in mind, that there still might be some kind of step-size restriction.
The global truncation error $\|u(t_n) -u_n\|$ is just the accumulation of the (local) truncation error.
The important part is, that you consider $h→0$. If we are speaking about convergence, we mean convergence with respect to the step-size $h$ getting smaller. 
Therefore, you can't tell if your RK method is of order $4$, if you only calculate the result for one step-size $h$.  
Assuming convergence of your problem, try the following: 
Solve for $h_i=2^{-i}$, $i=0,1,2,...$ and calculate:
$$α_i = \frac{1}{\log(2)} \log\left(\left|\frac{a(h_i)-a(h_{i+1})}{a(h_{i+1}) - a(h_{i+2})}\right|\right)$$
with $a(h) = u(t_{N})$ for stepsize $h$.
You should see $α_i\approx 4$.
Do a check and solve with some Euler method or Crank Nicolson.
The formula above assumes, that there exists some convergence $a(h)-a = \mathcal{O}(h^α)$. 
