I have the recursive Equation $$ T(n) = \begin{cases} 0 & \text{for } n = 0,\\ 1 & \text{for } 0 < n \leq 2,\\ \displaystyle 1 + \frac{2}{n} \sum_{i=1}^{n-2} T(i) &\text{else.} \end{cases} $$
Now I want to solve it for large $n$.
Experimentally, I could determine $\lim_{n \to \infty} T(n)/n \approx 0.4324314$, but is there a systematic proof for that?
// EDIT: Experimental result corrected. In my calculation, one constant had slipped up into the nominator, sorry.
I suggest you to consider the generating function $f(t) = \sum_{n=0}^{\infty}T(n)t^n$. Then, one can prove by the definition of $T(n)$ that $$\frac{\mathrm{d}}{\mathrm{d}t}f(t) = \frac{2t}{1-t} f(t) + \frac{1}{(1-t)^2}$$ with the initial condition $f(0)=0$.
This differential equation has a solution $$f(t) = \frac{1-e^{-2t}}{2(1-t)^2} = \mbox{tedious computations}= \sum_{n=1}^{\infty} \left( \sum_{k=1}^n \frac{(n+1-k)}{k!}(-2)^{k-1} \right) t^n$$ so that $$T(n)=\sum_{k=1}^n \frac{(n+1-k)}{k!}(-2)^{k-1}$$
PS: Thanks to Christian Blatter who patiently checked and corrected my computations. If one considers large $n$ one can approximate $T(n)$ with
$$\sum_{k=1}^{\infty} \frac{(n+1-k)}{k!}(-2)^{k-1} = \frac{n+1}{-2} \sum_{k=1}^{\infty} \frac{1}{k!}(-2)^{k} - \sum_{k=1}^{\infty} \frac{k}{k!}(-2)^{k-1} = \frac{1-e^{-2}}{2} (n+1) - e^{-2}$$ so that $$\lim_{n \to + \infty} \frac{T(n)}{n} = \frac{1-e^{-2}}{2}$$