I have a question that states:
Let $f(n)$ and $g(n)$ be positive non-decreasing functions such that $(f(n)−g(n)) = Θ(1)$. Prove that $f(n) = Θ(g(n))$.
The question's solution is one that I do not understand:
By definition, there exists $c_1, c_2 > 0$ such that for $n > n_0$ we have $$c_1 \leq f(n) - g(n) \leq c_2$$.
This shows that for $n > n_0$ we have: $$g(n) \leq c_1 + g(n) \leq f(n)$$.
Since $g(n) > 0$ and $g(n)$ is non-decreasing, we get that for $n > n_0$ we have: $$\frac{f(n)}{g(n)} - 1 \leq \frac{c_2}{g(n)} \leq \frac{c_2}{g(1)}$$
Given this, it follows that: $$f(n) \leq (\frac{c_2}{g(1)} + 1) \cdot g(n)$$
I don't really understand the proof here, and I especially don't understand how this proves that $f(n) = \Theta(g(n))$, because if you take $f(n) = 1$ and $g(n) = \frac{1}{n}$, this will give you $f(n) \neq \Theta(g(n))$. Can anyone help me understand?
Edit: I see now that the "counterexample" I provided is not a non-decreasing function. I guess it makes sense now...
Thank you.