The condition $n\log n \,a_n\to 0$ does not seem to be necessary.
Let us try to find a counterexample of the form
$$a_n=\frac{c_n}{n\log n}\quad,\quad n\geq 2.$$
The sequence $(a_n)$ will be decreasing if and only if $\frac{c_{n+1}}{c_n}\leq \frac{(n+1)\log(n+1)}{n\log n}$ for all $n$; and for that it is enough to have
$$c_{n+1}\leq \frac{n+1}n\, c_n\, .$$
It is also required that
$$\limsup c_n>0\qquad{\rm and}\qquad\sum_{n\geq 2} \frac{c_n}{n\log n}<\infty\, .$$
Set $n_1:=2$, and let $(n_k)_{k\geq 1}$ be a fast increasing sequence of integers, fast enough to have
$$ \frac{n_{k+1}}{n_k}\uparrow\infty\qquad{\rm and}\qquad\sum_{k=1}^\infty \frac{1}{\log(n_k)}<\infty\, .$$
Now, define a sequence $(c_n)$ as follows:
$$c_{n}=\frac{n}{n_{k+1}}\qquad{\rm if}\qquad n_k\leq n<n_{k+1}\cdot$$
Then $\limsup c_n\geq 1$ because $c_{n_{k+1}-1}=\frac{n_{k+1}-1}{n_{k+1}}\to 1$ as $k\to\infty$. Also, $c_{n+1}\leq \frac{n+1}nc_n$ for all $n$. Finally,
\begin{eqnarray}
\sum_{n=2}^\infty \frac{c_n}{n\log n}&=&\sum_{k=1}^\infty\frac1{n_{k+1}}\sum_{n_k\leq n<n_{k+1}}\frac{1}{\log n}\\
&\leq&\sum_{k=1}^\infty\frac1{n_{k+1}}\times \frac{n_{k+1}-n_k}{\log(n_k)}\\
&\leq&\sum_{k=1}^\infty \frac{1}{\log(n_k)}<\infty\, .
\end{eqnarray}
There is nothing special with $\log n$ here. The same proof shows that for any increasing function $\phi:\mathbb N\to\mathbb R^+$ tending to $\infty$, one can find a decreasing sequence $(a_n)$ such that $\sum a_n$ is convergent and $n\phi(n)\, a_n\not\to 0$.
The leftmost part of the inequality $|b_n|-|a_n|\leq|b_n-a_n|<\epsilon$ isn't needed in the proof.
Claim 1 $(a_n)$ is bounded above.
Proof Let $\epsilon>0$. From $|b_{n+1}-a_{n+1}|<\epsilon$, $a_{n+1}<b_{n+1}+\epsilon$. Use the monotonicity of $(a_n)$ and $(b_n)$. We have
$$a_1 \le \dots \le a_n\le a_{n+1} < b_{n+1}+\epsilon \le b_n+\epsilon \le \dots \le b_1 + \epsilon.$$
Since the choice of $n$ in the above inequality is arbitrary, we have $a_n < b_1 + \epsilon$ for all $n \in \Bbb N$.
Therefore, $(a_n)$ is bounded above by $b_1 + \epsilon$.
Similarly, we have another claim.
Claim 2 $(b_n)$ is bounded below.
Now, recall that $(a_n)$ and $(b_n)$ are increasing and decreasing sequences respectively, and apply MCT to $(a_n)$ and $(b_n)$ to establish the existence of $\lim a_n$ and $\lim b_n$. Finally, use $\lim\limits_{n\to+\infty}(b_n-a_n)=0$ to conclude that $\lim a_n = \lim b_n$.
Sorry for using others' ideas in my solution. I would like to draw a commutative diagrams in the comments, but the system forbids me from posting comments with two or more @ characters, so I can't post the following diagram in a comment. Hoping that others can benefit from his answer at the first glance, I draw this diagram for fun.
A graphical explanation to DonAntonio's answer
$\require{AMScd}$
\begin{CD}
@. a_n \\
@. @AA \vdots A \\
@. a_{N+1} \\
@. @AA (a_n)\uparrow A \\
@. a_N \\
\text{Suppose }a_N > b_K. \\
b_K @. \\
@V(b_n)\downarrow VV @.\\
b_{K+1} @. \\
@V \vdots VV @.\\
b_n @.
\end{CD}
It's clear from the diagram that we have to take $n \ge \max\{K,N\}$.
But $\lim\limits_{n\to+\infty}(b_n-a_n)=0$, contradiction.
Best Answer
The first statement says that $a_{n+1}/a_n \le \lambda$ but $$ \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lambda \in (1,\infty). $$ So if $1 < \lambda_0 < \lambda$, eventually (for $n$ large enough), we will have $$ \frac{a_{n+1}}{a_n} > \lambda_0 \iff a_{n+1} > \lambda_0 a_n. $$
Here, $$ \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lambda \in [0,1) $$ so analogously letting $0 \le \lambda \le \mu < 1$, for $n$ large enough you will get $a_{n+1} < \mu a_n$, and from that $n$ onwards, we can bound $\sum a_n$ by a geometric series.
UPDATE
Proof
Since $\lim_n b_n \to L$, there must exist $N \in \mathbb{N}$ such that $\forall \epsilon >0$ we have $|b_n - L| < \epsilon$ whenever $n > N$. In particular, pick $\epsilon = (L- \ell)/2$. Hence, there must exists $N \in \mathbb{N}$ such that for all $n > N$ we have $|b_n - L| < \epsilon$. Since $b_n < L$, we have $|b_n - L| = L-b_n$ and our inequality becomes $$L -b_n < \frac{L - \ell}{2}.$$ Eliminating $L/2$ and solving for $b_n$ on the LHS we get $$ b_n > L - \frac{L - \ell}{2} = \frac{L + \ell}{2} > \frac{\ell + \ell}{2} = \ell. $$