Asymptotic of an interesting recurrence relation I want to study the asymptotic behavior of the following recurrence relation:
$y_1=1$;
$y_{n+1}=y_n+\left(1+\frac{y_n}{n}\right)^{-n}$ for $n\ge 1$.
I made an initial attempt and guessed that $y_n∼Log(n)$. Any referrence or hint about this might be helpful. Any help will be greatly appreciated.
 A: Let $a_n=\frac{y_n}{\log{n}}$ when $n\ge 2$. Now we will show by induction that $a_n> 1$. This is because $a_2=3/(2\log{2})>1.$ Suppose $a_n>1$, we will show that
$a_{n+1}>1$. This is because the function
$$f_n(x)=\frac{1}{\log{(n+1)}}\left\{x\log{n}+\left(1+\frac{\log{n}}{n}x\right)^{-n}\right\}$$
is increasing when $x>0$(by computing its derivative.)
Now $$a_{n+1}=f_n(a_n)>f_n(1)=\frac{1}{\log{(n+1)}}\left\{\log{n}+\left(1+\frac{\log{n}}{n}\right)^{-n}\right\}.$$
To show $a_{n+1}>1$ it suffices to show
$$\left(1+\frac{\log{n}}{n}\right)^{-n}>\log\left(1+\frac{1}{n}\right).$$
Note when $x>0$ the inequality holds:
$$\log(1+x)<x.$$
Thus
$$\left(1+\frac{\log{n}}{n}\right)^{-n}=\exp\left(-n\log\left(1+\frac{\log{n}}{n}\right)\right)>\exp\left(-n\frac{\log{n}}{n}\right)=\frac{1}{n}>\log\left(1+\frac{1}{n}\right).$$
Once again, we may use similar technique to show that $b_n=\frac{y_n}{\log(n)+1}<1$.

Addendum Let $g_n(x)=\frac{1}{\log(n+1)+1}\left\{x(\log n+1)+\left(1+\frac{\log{n}+1}{n}x\right)^{-n}\right\}.$
Also, we see that 
$$b_{n+1}=g_n(b_n)<g_n(1)=\frac{1}{\log(n+1)+1}\left\{(\log n+1)+\left(1+\frac{\log{n}+1}{n}x\right)^{-n}\right\}.$$
It suffices to prove that
$$\left(1+\frac{\log{n}+1}{n}x\right)^{-n}=\exp\left(-n\log\left(1+\frac{\log{n}+1}{n}\right)\right)< \log\left(1+\frac{1}{n}\right).$$
Consider the inequality $\log(1+x)>x-x^2/2$. We see that
$$\exp\left(-n\log\left(1+\frac{\log{n}+1}{n}\right)\right)<\exp\left(-n\left(\frac{\log{n}+1}{n}-\frac{(\log{n}+1)^2}{2n^2}\right)\right)=\frac{1}{en}\exp\left(\frac{(\log{n}+1)^2}{2n}\right)<\frac{1}{n}-\frac{1}{2n^2}$$
The last inequality is because by taking derivative, and note that $n$ is integer
$$\frac{(\log{n}+1)^2}{2n}\le \frac{(\log 3+1)^2}{6}.$$
Also, this implies
$$\exp \frac{(\log{n}+1)^2}{2n} \le \exp \frac{(\log 3+1)^2}{6}<2.1 $$
Thus, we only need to prove 
$$\frac{2.1}{en}<\frac{1}{n}-\frac{1}{2n^2}$$
which holds when $n\ge 3$.
But we can check that $b_1,b_2,b_3$ all less than $1$. Thus we are done. 
