If $a_n+\log a_n = 1+\dfrac{1}{n}$, compute:
If $a_n$ is convergent to $l$, then passing to limit $l+\log l = 1$. But $f(x)=x+\log x$ is increasing, so unique solution is $l = 1$. But I don't know how to prove $a_n$ is bounded and monotonic.
Well, if you've already seen that $f:(0,\infty) \to \mathbb{R},\ f(x) = x+\ln x$ is increasing, then it's pretty clear that the sequence is decreasing, because:
$$f(a_n)=1+\frac{1}{n}>1+\frac{1}{n+1}=f(a_{n+1})$$
However, I think it's easier if you note that $f$ is continuous and increasing and that:
$$f(1)=1+\ln 1=1 < 1+\frac{1}{n}$$
$$f\left(1+\frac{1}{n}\right)=1+\frac{1}{n}+\ln\left(1+\frac{1}{n}\right) > 1+\frac{1}{n}$$
so there exists a unique real number $a_n \in\left(1,1+\frac{1}{n}\right)$ such that
$$f(a_n) = 1+\frac{1}{n}$$
Now, squeezing, we obviously get $\lim\limits_{n\to \infty} a_n=1$.
For the second limit, let $b_n:=a_n-1\to 0$ and write the formula as:
$$nb_n\left(1+\frac{\ln(1+b_n)}{b_n}\right)=1$$
Now, just passing to limit and using the well-know:
$$\lim_{x\to 0} \frac{\ln(1+x)}{x}=1$$
we get
$$\lim_{n\to \infty}nb_n=\frac{1}{2}$$
First part:
$\log a_ne^{a_n}=1+1/n$;
$z_n:=a_ne^{a_n}>e$;
$\log z_n$ is decreasing $\rightarrow$
$z_n$ is decreasing $\rightarrow$ $a_n$ is decreasing.
(Note: $f(x)=xe^x$ is increasing)
Bounded below:
$a_n>e/e^{a_n}>0$.
Hence convergent.
Only solution to $ae^a =e$ is $a=1$.
