Constant related to $f(n) = f(n-1) + \frac{1}{n f(n-1)}$

Question

Consider the sequence $f$ :

$$f(1) = 1$$

$$ f(n) = f(n-1) + \frac{1}{n f(n-1)}$$

Now we have for large $n$ :

$$ f(n) = \sqrt {2 \ln(n)} + \frac{1}{19} + C + eps(n) $$

Where $C$ is a constant and $eps(n)$ Goes to $0$ for large $n$.

Main question : Is $C = 0 $ ? If not What is Its value and do we have a closed form ??

Second question : How does $eps$ behave ? Do we have a closed form ?

Answer 1

You get (using $\ln(1+x)<x$ for $x>0$) $$ f(n)^2> f(n-1)^2+\frac2n\ge f(n-1)^2+2\ln(1+\frac1n) $$ so that $$ f(n)^2-2\ln(n+1)>f(n-1)^2-2\ln(n)>...>f(1)^2-2\ln(2)=1-2\ln(2) $$ Now use that growth to find a bound on the error in the first inequality and work toward the claim.

After thus establishing $f(n)\approx \sqrt{2\ln(n)}$ this can be re-inserted into the original equation \begin{align} f(n)^2&=f(n-1)^2+\frac2n+\frac1{n^2f(n-1)^2} =1+2\sum_{k=2}^n\frac1k+\sum_{k=2}^n\frac1{k^2f(k-1)^2} \\&\approx 2\ln(n)+c+O\left(\frac1{n}\right) \end{align} so that in the square root, using $\sqrt{1+x}=1+x/2+O(x^2)$ $$ f(n)=\sqrt{2\ln(n)}+\frac{c}{2\sqrt{2\ln(n)}}+O\left(\frac1{\ln(n)^{3/2}}\right) $$