To be clear, I am trying to find a closed form estimation for the $f(n)$, if $f(0)$ is defined to be $1$.
For example, this sequence would go as so: 1 is the starting value, and then add ${1\over1}$ to get the next value. $1+{1\over1}=2$, so the next value is 2. Iterating here, $2+{1\over2}=2.5$, so $f(2)=2.5$. $f(3)={2.5+{1\over2.5}}=2.9$. $f(4)={2.9+{1\over2.9}}=3.2448...$
I am wondering whether there exists a closed-form expression that can either exactly output or accurately estimate $f(n)$.
So far, an interesting relation between $f(n)$ and $n$ seems to be that as $n$ increases, the number $a$ such that $f(n)^a=n$ approaches a number close to 2. This was tested with Mathematica, and $f(100,000,000)^{1.927471...}=100,000,000$. It could also be the case that the value $a$ diverges very slowly, but it is difficult to check for higher numbers as Mathematica is having trouble witht the calculations.
By assuming the value of $a$ was actually approaching 2, and playing around with the graph on desmos, I managed to come up with the expression $\log_{(2-{1\over n+1})}{(n+1)}+{n+1 \over 10.7}$. Or, using the change of base formula to avoid the multi-term log base, ${\log{(n+1)} \over \log{(2-{1\over n+1})}}+ {n+1 \over 10.7}$. This expression seems to accurately estimate f(n) for the first 30 terms or so, but it soon becomes inaccurate for larger terms.
Is there any better formula that can accurately calculate the terms no matter how large? Or, is there some closed-form expression that is asymptotic with this function?
