A convergence test for series I need to prove the following result: let $\left( a_{n}\right) \subset
\left( 0,\infty \right) $ such that
\begin{equation*}
\frac{a_{n+1}}{a_{n}}=1-\frac{1}{n}-\frac{x_{n}}{n\ln n},
\quad
(X)
\end{equation*}
where $x_{n}\geq x>1$. Prove that
\begin{equation*}
\sum_{n\geq 1}a_{n}
\end{equation*}
is convergent. In addition, if
\begin{equation*}
\frac{a_{n+1}}{a_{n}}=1-\frac{1}{n}-\frac{x_{n}}{n\ln n},
\end{equation*}
where $x_{n}\leq x<1$, then
\begin{equation*}
\sum_{n\geq 1}a_{n}
\end{equation*}
is divergent.
For the first asertion it is clear. More precisely, the condition (X) can be
rewritten as follows
\begin{equation*}
a_{n}\left[ \left( n-1\right) \ln n-1\right] -a_{n+1}n\ln n=\left(
x_{n}-1\right) a_{n}.
\end{equation*}
Since $x_{n}\geq x>1$, then $x_{n}-1\geq x-1>0$
\begin{equation*}
a_{n}\left[ \left( n-1\right) \ln n-1\right] -a_{n+1}n\ln n\geq \left(
x-1\right) a_{n},
\end{equation*}
and, using the inequality
\begin{equation*}
\left( n-1\right) \ln \left( n-1\right) >\left( n-1\right) \ln n-1,
\end{equation*}
we obtain that
\begin{equation*}
a_{n}\left( n-1\right) \ln \left( n-1\right) -a_{n+1}n\ln n\geq \left(
x-1\right) a_{n}>0.
\end{equation*}
Hence $u_{n}=a_{n}\left( n-1\right) \ln \left( n-1\right) ,n\geq 2$, is
strictly decreasing, so it converges. Whence the telescoping sum associated
with $\left( u_{n}\right) $, that is
\begin{equation*}
\sum_{n\geq 1}\left( u_{n}-u_{n+1}\right) ,
\end{equation*}
is convergent. But
\begin{equation*}
0<\left( x-1\right) a_{n}\leq u_{n}-u_{n+1},
\end{equation*}
for all $n\geq 2$. By comparison test, it follows that the series
\begin{equation*}
\sum_{n\geq 1}a_{n}
\end{equation*}
is convergent.
Now the problem is the second assertion. Since
\begin{equation*}
a_{n}\left[ \left( n-1\right) \ln n-1\right] -a_{n+1}n\ln n=\left(
x_{n}-1\right) a_{n}.
\end{equation*}
and since $x_{n}\leq x<1$, then $x_{n}-1\leq x-1<0$,
\begin{equation*}
a_{n}\left[ \left( n-1\right) \ln n-1\right] -a_{n+1}n\ln n\leq \left(
x-1\right) a_{n}<0,
\end{equation*}
If one can guarantee that
\begin{equation*}
a_{n}\left( n-1\right) \ln \left( n-1\right) -a_{n+1}n\ln n\leq \left(
x-1\right) a_{n}<0,
\quad(XX)
\end{equation*}
for all $n\geq 2$, then it follows that $\left( u_{n}\right) $ is
increasing. So
\begin{equation*}
u_{n}=a_{n}\left( n-1\right) \ln \left( n-1\right) >\alpha ,
\end{equation*}
for all $n\geq 2$ and some $\alpha \in \left( 0,\infty \right) $. Hence, for
all $n\geq 2$, we have
\begin{equation*}
\frac{\alpha }{\left( n-1\right) \ln \left( n-1\right) }<a_{n}
\end{equation*}
and, since $\sum \frac{1}{\left( n-1\right) \ln \left( n-1\right) }$ is
divergent, it follows that the series $\sum a_{n}$ is divergent. How can I
prove that (XX), if it is correct of course. If not, how can we tackle the
proof of the second assertion?

Here is a method which can be applied to both cases : let's do the first one, the other one will be similar. We have $a_n > 0$ such that
$$\frac{a_{n+1}}{a_n} = 1 - \frac{1}{n} - \frac{x_n}{n \ln (n)}$$
with $x_n \geq x > 1$ for a real number $x$. Let $y=(1+x)/2$. Consider $$u_n = \frac{1}{n \ln^{y}(n)}$$
One has
$$\frac{u_{n+1}}{u_n} = \frac{n \ln^y(n)}{(n+1)\ln^y(n+1)} = \left(\left(1+ \frac{1}{n} \right)\left( \frac{\ln(n+1)}{\ln(n)}\right)^y \right)^{-1}$$ $$ =\left(\left(1+ \frac{1}{n} \right)\left( 1+\frac{\ln(1+\frac{1}{n})}{\ln(n)}\right)^y \right)^{-1}=\left(\left(1+ \frac{1}{n} \right)\left( 1+\frac{1}{n\ln(n)}+o\left(\frac{1}{n\ln(n)} \right)\right)^y \right)^{-1}$$
$$=\left(\left(1+ \frac{1}{n} \right)\left( 1+\frac{y}{n\ln(n)}+o\left(\frac{1}{n\ln(n)} \right)\right) \right)^{-1} =\left(1+ \frac{1}{n} +\frac{y}{n\ln(n)}+o\left(\frac{1}{n\ln(n)} \right) \right)^{-1} $$ $$=1- \frac{1}{n} -\frac{y}{n\ln(n)}+o\left(\frac{1}{n\ln(n)} \right) $$
Now, because $$x_n - y \geq  \frac{x-1}{2} > 0$$
you see that from a certain rank, you will have
$$\frac{u_{n+1}}{u_n} \geq \frac{a_{n+1}}{a_n}$$
Because the series $\sum u_n$ is convergent, you deduce that $\sum a_n$ is also convergent by comparison.
