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
  $$a_{n+1}<a_{n}$$
  $$a_{n}>\ln{\left(\dfrac{1}{n\ln{\left(1+\frac{1}{n}\right)}}\right)}?$$

 A: 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, $a_{n+1}\lt a_n$.
A: I can prove the second inequality by induction and calculus.
Base case. $$a_{1}=1=-\ln(\ln(e))>-\ln(\ln(2))=\ln{\left(\dfrac{1}{\ln{\left(2\right)}}\right)}$$
Induction hypothesis. Assume that:
$$a_{n}>\ln{\left(\dfrac{1}{n\ln{\left(1+\frac{1}{n}\right)}}\right)}$$

Lemma 1. For all $k>1$ and $x>0$, $f(x)=x+\frac{1}{e^{x}k}$ is increasing.
Proof. $$f'(x) = 1-\frac{1}{e^{x}k}$$
For $x>0$ we have that $e^{x}k>e^{-\ln(k)}k=1$ thus $f'(x)>0$.

Induction step.
\begin{align*} a_{n+1} &= a_{n}+\ln{\left(\dfrac{n}{n+1}\right)}+\dfrac{1}{e^{a_{n}}\cdot n} \\ &> \ln{\left(\dfrac{1}{n\ln{\left(1+\frac{1}{n}\right)}}\right)} + \ln{\left(\dfrac{n}{n+1}\right)}+\dfrac{1}{e^{\ln{\left(\dfrac{1}{n\ln{\left(1+\frac{1}{n}\right)}}\right)}}\cdot n} \\&= \ln{\left(\dfrac{1}{n\ln{\left(1+\frac{1}{n}\right)}}\right)} + \ln{\left(\dfrac{n}{n+1}\right)}+\dfrac{1}{\dfrac{1}{n\ln{\left(1+\frac{1}{n}\right)}}\cdot n} \\&= \ln{\left(\dfrac{1}{n\ln{\left(1+\frac{1}{n}\right)}}\right)} + \ln{\left(\dfrac{n}{n+1}\right)}+\ln{\left(1+\frac{1}{n}\right)} \\&= \ln{\left(\dfrac{1}{n\ln{\left(1+\frac{1}{n}\right)}}\right)} \\&> \ln{\left(\dfrac{1}{(n+1)\ln{\left(1+\frac{1}{(n+1)}\right)}}\right)}   \end{align*}
We may do the second step because of the IH and Lemma 1.
The last step is also fairly straightforward. It is a well known fact that $\left(1+\frac{1}{n}\right)^n$ is increasing. Thus $\ln\left(\left(1+\frac{1}{n}\right)^n\right)$ is increasing. Thus $\frac{1}{\ln\left(\left(1+\frac{1}{n}\right)^n\right)}$ is decreasing and so is the logarithm of that.
This completes the proof.
A: Update: Not an Answer. Since the question was edited, this is no longer a valid answer. It is only slightly useful. It just shows by basic algebraic manipulations that if the second inequality holds, then the sequence monotonically decreases. I won't take down this answer though, as I feel at least it adds some more depth to the question i.e. it shows that the second condition implies the first. However, it does not "crack" the problem with any ingenious insight or idea, and definitely does not deserve the bounty for this question. I hope this update encourages others to attempt a real answer to this question.

To start, let's introduce some notation to tidy things up. Let:
$$b_n = \left(\dfrac{1}{n\ln{\left(1+\frac{1}{n}\right)}}\right)$$
Now we notice that the second condition is just $a_n > \ln{b_n}$. We show that this implies $a_{n + 1} < a_n$. Proceed as follows. Assuming $a_n > \ln{b_n}$ then:
$$\dfrac{1}{e^{a_{n}}\cdot n} < \dfrac{1}{e^{\ln{b_{n}}}\cdot n} = \dfrac{1}{b_{n}\cdot n} = \ln{\left(1 + \frac1{n}\right)} = \ln{\left(\frac{n + 1}{n}\right)}$$
Now plugging in the recurrence in the question for $a_{n + 1}$ we get:
$$a_{n+1}=a_{n}+\ln{\left(\dfrac{n}{n+1}\right)}+\dfrac{1}{e^{a_{n}}\cdot n} < \\
a_{n}+\ln{\left(\dfrac{n}{n+1}\right)}+\ln{\left(\frac{n + 1}{n}\right)} = a_n + \ln{\left(\frac{n}{n + 1}\cdot\frac{n + 1}{n}\right)} = a_n + \ln{1} = a_n$$
So $a_{n + 1} < a_n$ as required.
