<s>[INST] A Problem Concerning the Series Problem
Let $\{a_n\},\{b_n\}$ be two sequences such that $a_n>0(\forall n \geq 1)$, $\sum\limits_{n=1}^{\infty}b_n$ is absolutely convergent, and $$\frac{a_n}{a_{n+1}}\leq 1+\frac{1}{n}+\frac{1}{n\ln n}+b_n(\forall n \geq 2).\tag{1}$$
Prove

*

*$\dfrac{a_n}{a_{n+1}}< \dfrac{n+1}{n}\cdot \dfrac{\ln(n+1)}{\ln
    n}+b_n(\forall n \geq 2)\tag{2};$

*$\sum\limits_{n=1}^{\infty}a_n$ is divergent.

Proof for 1
It suffices to show
\begin{align*}
&~~~1+\frac{1}{n}+\frac{1}{n\ln n}<\dfrac{n+1}{n}\cdot \dfrac{\ln(n+1)}{\ln
    n}\tag{3}\\
\Leftrightarrow &~~~n\ln n+\ln n+1< (n+1)\ln(n+1)\\
\Leftrightarrow &~~~\ln n^{n+1}+1< \ln(n+1)^{n+1}\\
\Leftrightarrow &~~~\ln\left(1+\frac{1}{n}\right)^{n+1}> 1\\
\Leftrightarrow &~~~\left(1+\frac{1}{n}\right)^{n+1}> e.\tag{4}
\end{align*}
Define
$$x_n:=\left(1+\frac{1}{n}\right)^{n+1},n=1,2,\cdots$$
By AM-GM inequality, we obtain
\begin{align*}
\frac{1}{x_n}&=\frac{1}{\left(1+\dfrac{1}{n}\right)^{n+1}}=\left(1-\frac{1}{n+1}\right)^{n+1}\\
&=\left(1-\frac{1}{n+1}\right)\cdot\left(1-\frac{1}{n+1}\right)\cdots\left(1-\frac{1}{n+1}\right)\cdot 1\\
&< \left[\dfrac{\left(1-\dfrac{1}{n+1}\right)+\left(1-\dfrac{1}{n+1}\right)+\cdots+\left(1-\dfrac{1}{n+1}\right)+1}{n+2}\right]^{n+2}\\
&= \left[\dfrac{(n+1)\left(1-\dfrac{1}{n+1}\right)+1}{n+2}\right]^{n+2}\\
&=\frac{1}{\left(1+\dfrac{1}{n+1}\right)^{n+2}}\\
&=\frac{1}{x_{n+1}},
\end{align*}
which implies $x_n > x_{n+1}$, namely $\{x_n\}$ is a decreasing sequence. Moreover, we can find
$$\lim_{n \to \infty}x_n=e.$$
Hence
$$\inf_{x \in \mathbb{N^+}} x_n=e,$$
which implies
$$x_n>e,$$which is desired.
AM I RIGHT? HOW TO TACKLE 2?
 [/INST] Step 1 is correct though the simple estimate $\log{(1+\frac{1}{n})} > \frac{1}{n} - \frac{1}{2n^2} \ge \frac{1}{n+1}$ immediately shows (4)
Now let $x>A>0, |y|<\frac{x}{2}$, then $\frac{1}{x+y}=\frac{1}{x}+z, |z| < \frac{2|y|}{A^2}$ since obviosuly $z=-\frac{y}{x(x+y)}$ and $x(x+y)>\frac{x^2}{2}> \frac{A^2}{2}$ by our assumptions.
Let $c_n=\frac{1}{n\ln n}, n \ge 2, \frac{c_n}{c_{n+1}}>1, \Sigma c_n=\infty$ 
Using the absolute convergence of $b_n$ we can find $N$ s.t. $|b_n|<\frac{1}{8}, n>N$, so by the above we can write $\frac{1}{\frac{c_n}{c_{n+1}}+b_n}=\frac{c_{n+1}}{c_n}+d_n, |d_n|<2|b_n|<\frac{1}{4}, \Sigma{|d_n|}=B<\infty$,all for $n>N$ of course.
In particular relation (2) becomes:
$\frac{a_{n+1}}{a_n}> \frac{c_{n+1}}{c_n}+d_n, n>N$ 
Taking logarithms we get
$\log a_{n+1} - \log a_{n}>\log c_{n+1} - \log c_{n}+\log(1+\frac{c_n}{c_{n+1}}d_n) \ge \log c_{n+1} - \log c_{n} - 2|d_n|, n>N$, 
using $\frac{c_n}{c_{n+1}}|d_n| < 2|d_n| < \frac{1}{2}, \log{(1+x)}\ge -2|x|, 0 \le |x|<\frac{1}{2}$ 
summing now from say $N+1$ to $N+p$, we get:
$\log a_{N+p} - \log a_{N+1} > \log c_{N+p} - \log c_{N+1} -2B$ or 
$a_{N+p}>(\frac{a_{N+1}}{c_{N+1}e^{2B}})c_{N+p}=Kc_{N+p}, p\ge 1$. 
Combining with the divergence of $\Sigma{c_n}$ we are done!
Note that we proved the following result:
$a_n>0, \frac{a_n}{a_{n+1}}< \frac{c_n}{c_{n+1}}+b_n$, where $c_n>0, \Sigma{c_n}=\infty, \Sigma{|b_n|}<\infty, 0<C_1<\frac{c_n}{c_{n+1}}<C_2$ implies $\Sigma{a_n}=\infty$, where here we have $C_1=1, C_2=2$ but obviously we can easily adjust the proof for general $C_1, C_2$
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