I'm working on understanding Nagura's analysis of the upper bound for $\psi(x)$ which is done in Lemma 2. I am unclear on one step of his reasoning.
With Lemma 1, he establishes for $x \ge 2000$:
$$\log\Gamma(\lfloor{x}\rfloor+1) - \log\Gamma(\lfloor{\frac{x}{2}}\rfloor+1) - \log\Gamma(\lfloor{\frac{x}{3}}\rfloor+1)- \log\Gamma(\lfloor{\frac{x}{7}}\rfloor+1) - \log\Gamma(\lfloor{\frac{x}{43}}\rfloor+1) - \log\Gamma(\lfloor{\frac{x}{1806}}\rfloor+1) < 1.0851x$$
With Lemma 2, he shows that for $x > 0$:
$$\log\Gamma(\lfloor{x}\rfloor+1) - \log\Gamma(\lfloor{\frac{x}{2}}\rfloor+1) - \log\Gamma(\lfloor{\frac{x}{3}}\rfloor+1)- \log\Gamma(\lfloor{\frac{x}{7}}\rfloor+1) - \log\Gamma(\lfloor{\frac{x}{43}}\rfloor+1) - \log\Gamma(\lfloor{\frac{x}{1806}}\rfloor+1) \ge \psi(x) -\psi(\frac{x}{1806})$$
Now, it's the next step that confuses me. Based on these two steps, Nagura makes the following argument:
$$\psi(x) < 1.0851(x + \frac{x}{1806} + \frac{x}{1806^2} + \frac{x}{1806^3} + \ldots ) < 1.086x$$
I am having trouble understanding this step.
Does it follow since he has established this step:
$$\psi(x) < 1.0851x + \psi(\frac{x}{1806})$$
And is trying to generalize to a statement of the form:
$$\psi(x) < Kx$$
Thanks,
-Larry