Prove that this sequence is decreasing for all $n$. 
Define $a_{1}=1$, and such 
  $$a_{n+1}=a_{n}+\ln{\left(\dfrac{n}{n+1}\right)}+\dfrac{1}{e^{a_{n}}\cdot n}$$
  show that
  $$a_{n+1}<a_{n}$$
  $$a_{n}>\ln{\left(\dfrac{1}{n\ln{\left(1+\frac{1}{n}\right)}}\right)}?$$


Let us prove that $a_{n+1}\lt a_n$.
First of all,
$$\begin{align}\\&a_{n+1}=a_n+\ln\left(\frac{n}{n+1}\right)+\frac{1}{e^{a_n}\cdot n}\\\\&\iff e^{a_{n+1}}=e^{a_n+\ln\left(\frac{n}{n+1}\right)+1/(e^{a_n}\cdot n)}\\\\&\iff e^{a_{n+1}}=e^{a_n}\cdot e^{\ln\left(\frac{n}{n+1}\right)}\cdot e^{1/(e^{a_n}\cdot n)}\\\\&\iff e^{a_{n+1}}=e^{a_n}\cdot \frac{n}{n+1}\cdot e^{1/(e^{a_n}\cdot n)}\\\\&\iff e^{a_{n+1}}(n+1)=e^{a_n}\cdot n\cdot e^{1/(e^{a_n}\cdot n)}\\\\&\iff b_{n+1}=b_n\cdot e^{1/b_n}\qquad (\text{set $b_n=e^{a_n}\cdot n$})\\\\&\iff \frac{1}{b_{n+1}}=\frac{1}{b_n}\cdot e^{-1/b_n}\\\\&\iff c_{n+1}=c_n\cdot e^{-{c_n}}\qquad (\text{set $c_n=\frac{1}{b_n}$})\\\\&\iff -c_{n+1}=-c_n\cdot e^{-c_n}\\\\&\iff d_{n+1}=d_n\cdot e^{d_n}\qquad (\text{set $d_n=-c_n$})\\\\&\iff e^{d_{n+1}}=\left(e^{d_n}\right)^{e^{d_n}}\\\\&\iff f_{n+1}={f_n}^{f_n}\end{align}$$
where $f_n=e^{d_n}=e^{-1/(ne^{a_n})},f_1=e^{-1/e}.$
Here, since $a_n=\ln\left(-\frac{1}{n\ln f_n}\right)$ and
$$\begin{align}a_{n+1}\lt a_n&\iff \ln\left(-\frac{1}{(n+1)\ln f_{n+1}}\right)\lt \ln\left(-\frac{1}{n\ln f_n}\right)\\\\&\iff -\frac{1}{(n+1)\ln f_{n+1}}\lt -\frac{1}{n\ln f_n}\\\\&\iff \frac{1}{(n+1)\ln f_{n+1}}\gt \frac{1}{n\ln f_n}\\\\&\iff n\ln f_n\gt (n+1)\ln f_{n+1}\\\\&\iff n\ln f_n\gt (n+1)f_n\ln f_n\\\\&\iff f_n\gt\frac{n}{n+1}\end{align}$$
it is sufficient to prove that $f_n\gt\frac{n}{n+1}$.
Using the fact that $$\text{$\left(\frac{n}{n+1}\right)^n\ \ \left(=\frac{1}{\left(1+\frac 1n\right)^n}\right)$ is a decreasing sequence}\tag1$$ let us prove that $f_n\gt\frac{n}{n+1}$ by induction on $n$.
For $n=1$, $f_1=e^{-1/e}\gt \frac 12$ because $2^e\gt 2^{2}= 4\gt e$.
Suppose that $f_n\gt\frac{n}{n+1}$. Then,
$$f_{n+1}={f_n}^{f_n}\gt \left(\frac{n}{n+1}\right)^{n/(n+1)}\gt \frac{n+1}{n+2}$$because of $(1)$.
So, we have $f_n\gt\frac{n}{n+1}$ for every $n$.
Thus, $a_{n+1}\lt a_n$.
