Trouble to understand an Analysis proof.

Question

I am currently studying the following proof, which may be found on this article (page 12):

However, I'm facing difficulties to proper understand some of the steps. Here are my questions:

1. Why is it possible to say that "there exists an integer $N$ and $\alpha > 0$ such that $\Sigma_n(x) \leq n^{1 - \alpha}$, for all $n \geq N$"?
2. How does integrating ${f_n}^{\prime} \geq n^{\alpha} f_n$ between $x^{\prime}$ and $x$ gives me that $f_n(x^{\prime}) \leq M e^{-\delta n^{\alpha}}$, for all $n \geq N$? Same question applies to the integration of ${f_n}^{\prime} \geq \frac{n}{\Sigma}f_n$ between $x^{\prime}$ and $x$.
3. Why does "$T_n(x)$ converge to $f(x)$ as $n$ tends to infinity"? And why does this imply that $f(x) - f(x^{\prime}) \geq (x - x^{\prime}) \left(\limsup_{n \rightarrow +\infty}\frac{\ln\Sigma_n(x^{\prime})}{\ln(n)}\right)$?

EDIT:

Question 1 remains unsolved.

Question 2 is easily answered by noticing that $\frac{f_n^{\prime}}{f_n} = \left(\ln f_n\right)^{\prime}$ for both cases.

The second part of question 3 may by answered by applying $\limsup$ on both sides of the given inequality; however, the first part remains unsolved. Why does $T_n(x)$ converge to $f(x)$ as $n$ tends to infinity?

Thanks in advance.

Answer 1

(1) $x < x_1$, so by defn of inf, $\limsup_{n \to \infty} \frac{\log\Sigma_n(x)}{\log n} < 1$, call the limsup $1-2\alpha$ for $\alpha > 0$. Then there is some $N$ so that for all $n \ge N$, $\frac{\log \Sigma_n(x)}{\log n} \le (1-2\alpha)+\alpha = 1-\alpha$. In other words, there is some $N$ so that for all $n \ge N$, $\Sigma_n(x) \le n^{1-\alpha}$.

(3) part 1: It suffices to prove the following general statement: Let $(a_n)_n$ be a sequence of real numbers such that $a_n \to a$. Then $\frac{1}{\log n}\sum_{i=1}^n \frac{a_i}{i} \to a$. Below is a hint.

$a_n \approx a$ for large $n$, so $\sum_{i=1}^n \frac{a_i}{i} \approx \sum_{i=1}^n \frac{a}{i} \approx a\log n$.