Limit of recursion relation

Question

Consider the sequence of functions $\{F_n\}_{n\in \mathbb{N}}$, where each $F_n$ is defined on $\{0,...,n\}$ by recurrence of the following form: $$ F_n(0)=3 \textrm{ and }F_n(k)=\frac{1}{k^2}+\frac{\left(1-\frac{1}{n}\right) (k+n+1)}{k+n+2} F_n(k-1), \textrm{ for } 0< k\leq n $$

Q: Is Mathematica able to prove that $$\lim_{n\rightarrow \infty} F_n(n)= \frac{18+\pi ^2}{12 e}=0.85438787439...$$

Motivation: I am developing a methodology to solve this type of problem and I would like to know if one can solve it in a simple way.

Answer 1

Mathematica can actually solve the recursion relation in closed form, $$F_n(n)=-\tfrac{1}{2}(n^2-1)^{-1}\left(\frac{n-1}{n}\right)^n\left[n \left(\frac{n}{n-1}\right)^n \Phi \left(\frac{n}{n-1},1,n+1\right)+n (n+2) \left(\frac{n}{n-1}\right)^n \Phi \left(\frac{n}{n-1},2,n+1\right)-(n-1) \left((n+2) \text{Li}_2\left(\frac{n}{n-1}\right)+3 n+\ln(1-n) +6\right)\right],$$ with $\Phi$ the Lerch transcendent and $\text{Li}_2$ the polylog. The large-$n$ limit then evaluates to $$\lim_{n\rightarrow\infty} F_n(n)=\frac{18+\pi^2}{12 e}.$$

Some generalisations:
If the initial value is varied, $F_n(0)=p$, the limit becomes $$\lim_{n\rightarrow\infty} F_n(n)=\frac{6p+\pi^2}{12 e}.$$ If the large-$n$ limit is evaluated at fixed $k$ Mathematica gives $$\lim_{n\rightarrow\infty} F_n(k)=p+\frac{\pi ^2}{6}-\psi ^{(1)}(k+1),\;\;n\gg k,$$ with $\psi^{(1)}$ the polygamma function.

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the Mathematica commands are:

result=f[k]/.RSolve[{f[k]==1/k^2+(1-1/n)*(k+n+1)*f[k-1]/(k+n+2),f[0]==3},f[k],k]/.k->n    

Limit[FullSimplify[Normal[Series[result,{n,Infinity,0}]],Assumptions->n>1],n->Infinity]