I've been asked to find an equivalent of the following recurrent sequence with $u_0=2$ : $u_{n+1}=u_n+\log u_n$.
It is clear that this is going to infinity with n, and I tried two things :
Solving the ODE $y'=\log y$, but I don't know what is the solution...
Finding a function $f$ such that $f(n+1,u_{n+1})-f(n,u_n)$ has a finite limit (non zero) in order to apply Cesaro.
But I'm stuck with only zeros limits, and numerically I observe that it's not growing much faster than $n^\alpha$...
Any clue about the sequence I have to consider ?
EDIT : You can also show that $u_{n+1}\sim u_n$