Construct a sequence Is it possible to construct a sequence $a_n$ satisfying
(1) $\forall n,a_n>0$
(2) $\limsup_{n\to \infty}\left(\frac{1+a_{n+1}}{a_n}\right)^n=e$
If I use the reccurance $a_n+1=(1+\frac{1}{n})a_n-1$ for all terms, the sequence will become negative after some steps, while if the reccurance holds for some term, the upper bound will be greater than $e$, so I'm confused whether the sequence exists or not.
Thank you for your attention!


$$a_n=1+n\log n\quad\text{or}\quad a_n=1+\sum\limits_{k=1}^n\log k$$

Then $a_{n+1}-a_n\sim\log n\sim\frac1na_n$ hence $\frac{1+a_{n+1}}{a_n}=1+\frac1n+o\left(\frac1n\right)$ and the LHS of (2) converges to $\mathrm e$.
Note that the somewhat more natural candidate $a_n\sim bn^c$ with $b\gt0$ and $c\geqslant1$ fails because then $a_{n+1}-a_n\sim\frac{c}na_n$ hence the LHS of (2) converges to $\mathrm e^c$ for every $c\gt1$, and to $\mathrm e^{b+1}$ if $c=1$, neither of which can be $\mathrm e$.
