Raabe's and Schlömilch's tests for limits This is problem 2.53 in Sohrab's Basic Real Analysis
Problem
First the Raabe's test as it has been formulated in the book.
Corollary 2.3.31 Let $(x_n)$ be a sequence of positive numbers. Then $\sum x_n$ converges if $x_{n+1}/x_n\leq 1-r/n$ is ultimately true for some $r>1$.
Problem 2.53
Show that Raabe's Test (Corollary 2.3.31) implies the following one (which is due to Schlömilch):
Let $x_n > 0\;\forall n\in\mathbb{N}$. Then $\sum x_n$ converges if $n\log(x_n/x_{n+1})\geq r$ is ultimately true for some $r > 1$. Hint: Use the inequalities $x/(1+x)\leq \log(1+x)\leq x$ for all $x>-1$.
Question
The idea is to prove: Schlömilch's premises $\implies$ Raabe's premises $\implies$ convergence. BUT I'm only able to prove: Raabe's premises $\implies$ Schlömilch's premises $\implies$ convergence:
$$\frac{r}{n}\leq 1-\frac{x_{n+1}}{x_n}\leq \log\frac{x_n}{x_{n+1}}$$
where, with $y:=\frac{x_{n}}{x_{n+1}}-1$, we have used $y/(1+y)\leq \log(1+y)$.
It seems to me that the Raabe's is a more general result, therefore it is not possible to show that Schlömilch's premises $\implies$ Raabe's premises.
And this is the question: Is my thought correct or otherwise where is the error in my reasoning?
 A: To deduce the premises of Raabe's test from the premises of Schlömilch's test, we need to take a different $r$ in Raabe's test than in Schlömilch's.
The key is that
$$\log \frac{1}{1-x} = \sum_{k = 1}^{\infty} \frac{x^k}{k} = x + O(x^2)$$
for $x > 0$, so given any $\varepsilon > 0$, there is a $\delta > 0$ such that
$$x < \log \frac{1}{1-x} < (1+\varepsilon)\cdot x\tag{1}$$
for all $0 < x < \delta$.
So let's choose a $\varepsilon > 0$ such that $1 < r' := \frac{r}{1+\varepsilon}$ and the corresponding $\delta > 0$. Choose $n_0$ so large that
$$\log \frac{x_n}{x_{n+1}} \geqslant \frac{r}{n}\tag{2}$$
for all $n \geqslant n_0$, and always let $n \geqslant n_0$ in the following. Now
$$\log \frac{x_n}{x_{n+1}} = \log \frac{1}{x_{n+1}/x_n} = \log \frac{1}{1 - \frac{x_n - x_{n+1}}{x_n}},$$
and when $\frac{x_n - x_{n+1}}{x_n} < \delta$ - we have $x_n > x_{n+1}$ by $(2)$ - we get
$$\frac{r}{n} \leqslant \log \frac{x_n}{x_{n+1}} < (1+\varepsilon)\frac{x_n-x_{n+1}}{x_n}\tag{3}$$
from $(1)$. But the inequality between the outer terms of $(3)$ is just
$$\frac{r'}{n} < 1 - \frac{x_{n+1}}{x_n},\tag{4}$$
so for these $n$, the premises of Raabe's test are satisfied with $r' > 1$. And if
$$\delta \leqslant \frac{x_n - x_{n+1}}{x_n} = 1 - \frac{x_{n+1}}{x_n},$$
then surely $(4)$ holds when $n$ is so large that
$$\frac{r'}{n} \leqslant \delta,$$
that is, for $n \geqslant \frac{r'}{\delta}$.
Hence, from
$$\frac{r}{n} \leqslant \log \frac{x_n}{x_{n+1}}$$
for all $n \geqslant n_0$ we deduce that
$$\frac{r'}{n} \leqslant 1 - \frac{x_{n+1}}{x_n}$$
for all $n \geqslant \max \{ n_0,\, r'/\delta\}$.
