Is this a valid conclusion from Chebyshev's first and second function I am reading through a paper on Arxiv.org.
I am confused by a step in Theorem 2.3.1 on page 39.
Kyle Balliet, the author of the paper, starts with standard definitions of the Chebyshev functions:


*

*$\psi(x) = \sum\limits_{m=1}^\infty\vartheta(x^{1/m})$ 

*$\vartheta(x)=\sum\limits_{p \le x \text{ and p is prime}}\log p$
His next step seems wrong.  
He writes:

Our proof conforms to S. Ramanujan’s [13] proof of Bertrand’s
  postulate. It also conforms with J. Nagura’s [12] proof of primes in
  the interval $n$ to $\frac{6n}{5}$.
The basis of our proof is to approximate the first Chebyshev function
  ϑ using the second Chebyshev function ψ

Here is Ramanujan's proof.
Here is the next step made by Balliet:


*

*$\psi(x) - \sum\limits_{p\le 503}\psi(x^{1/p})\ge \vartheta(x)$

*$\psi(x) - \sum\limits_{p\le 509}\psi(x^{1/p})\le \vartheta(x)$
I am not clear how this is true.  It does not appear to me that Balliet has proved this.  Am I wrong?  
If I am wrong, I would appreciate if someone could explain the argument that Balliet is making.

Edit 1: 
I contacted Kyle and he confirmed that his equation is based on the möbius inversion formula where:
$$\vartheta(x)=\sum\limits_{n=1}^{\infty}\mu(n)\psi(\sqrt[n]{x})$$
This suggests that it is a generalization of Jitsuro Nagura's first step in his theorem.
With this in mind, the reasoning may be a generalization of this.

Edit 2:
Unless I am missing something, it seems clear to me that (2.1) on page 40 has a typo and is wrong:

$$\psi(x) - \sum\limits_{p\le 503}\psi(x^{1/p})\ge \vartheta(x)$$

Jitsuro Naguro showed on page 181 of his proof:

$\vartheta(x) \ge \psi(x) - \psi(x^{1/2}) - \psi(x^{1/3}) - \psi(x^{1/5})$

It seems quite clear to me that:

$\psi(x) - \psi(x^{1/2}) - \psi(x^{1/3}) - \psi(x^{1/5})  > \psi(x) - \sum\limits_{2 \le p \le 5}\psi(x^{1/p}) >  \psi(x) - \sum\limits_{2 \le p \le 503}\psi(x^{1/p}) >  \psi(x) - \sum\limits_{1 \le p \le 503}\psi(x^{1/p})$ 

Am I wrong?
 A: You're not wrong - the proof is indeed flawed.  I'm very embarrassed as that error was actually printed and somehow slipped by the review committee.  With that said, however, the result still holds true that there is always a prime between $n$ and $\frac{520n}{519}$ for all $n\geq 31409$.  We can show this quite easily as:
Using Theorem 5.2 of P. Dusart, we can deduce for all $n \geq 3594641$:
$$\begin{align}
\vartheta\left(\tfrac{520n}{519}\right)-\vartheta(n) &> \left(\frac{520n}{519}-\frac{0.2\cdot\frac{520n}{519}}{\log^2 \frac{520n}{519}}\right)-\left(n+\frac{0.2n}{\log^2 n}\right)\\ &> \frac{n}{519}-\frac{2\cdot0.2\cdot\frac{520n}{519}}{\log^2 n}\\ &\stackrel{?}{>}0.
\end{align}$$
Now the last inequality holds if we are able to show $n>\frac{208n}{\log^2 n}$.  Dividing both sides by $n$ and applying a little algebra we find that for all $n>e^{4\sqrt{13}}\approx 1834358$ the inequality holds.
Therefore $\vartheta\left(\tfrac{520n}{519}\right)-\vartheta(n)>0$ for all $n \geq 3594641$ and we can use Mathematica or similar to verify the cases for $31409\leq n\leq 3594641$.
