# Confusion about Big O-Notation in convergence proof for differential algebraic equations

I understand almost all of this proof. But the part I can't find an explanation for is how they get the result for $$r$$ with the sums over the $$\mathcal{O}(h^{\nu})$$ all the way at the end of the proof. Can someone please explain this?

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My problem is the following:

They define the global error as $$\triangle z_{n+1} = \rho \triangle z_n + \delta_{n+1}$$ with $$\delta_{n+1} = \mathcal{O}{(h^{\min\lbrace p , q+1 \rbrace})}$$ and $$\rho = 1 - b^{T}A^{-1}\mathbb{1}$$ (where $$\rho$$ is the stability function of the runge kutta method) then after repeated use of this formula they get:

$$\triangle z_{n} = \sum \limits_{i=1}^{n} \rho^{n-i} \delta_i$$

and from this alone they follow

$$1) \triangle z_{n} = \mathcal{O}{(h^{\min\lbrace p , q+1 \rbrace})} \mbox{ for } -1 \leq \rho < 1$$

$$2) \triangle z_{n} = \mathcal{O}{(h^{\min\lbrace p-1 , q \rbrace})} \mbox{ for } \rho = 1$$

$$3) \mbox{The solution diverges} \mbox{ for } \rho > 1$$

And I don't understand how they get these answers.

Note that $$|\Delta z_n| \le C h^{\min\{p, q+1\}} \left|\sum_{i=1}^n \rho^{n-i}\right|.$$
• If $$|\rho| < 1$$, then the sum can be bounded by the geometric series $$1 + \rho + \rho^2 + \cdots = \frac{1}{1-\rho}$$.
• If $$\rho = -1$$, then the sum can be bounded by $$1$$.
• If $$\rho = 1$$, then the sum equals $$n \le C/h$$ (the relationship between $$n$$ and $$h$$ is an assumption stated in the theorem).