<s>[INST] Closed form of function $f(n) = \frac1n \sum\limits _{k=1}^{n-1} f(k)$? Could anyone help me get to the closed form of the function?  Here $\mathbb{N}=\{1,2,3,\ldots\}$.

Find all functions $f:\mathbb{N}\to\mathbb{R}$ such that $f(1)=1$ and
$$f(n) = \frac 1 n \sum _{k = 1}^{n-1}f(k)$$
for every integer $n>1$.

So far, I see that
$$f(2)=\frac{1}{2}f(1)=\frac12,$$
$$f(3)=\frac{1}{3}\left(f(1)+f(2)\right)=\frac{1}{3}\left(1+\frac{1}{2}\right)=\frac12,$$
$$f(4)=\frac14\left(f(1)+f(2)+f(3)\right)=\frac14\left(1+\frac12+\frac12\right)=\frac12.$$
I guess that $f(n)=\frac12$ for every $n\in\mathbb{N}$.  How to show this?
 [/INST] Start by computing the first few terms to gain an intuition into the problem, we can see that $f(2) = \frac{1}{2} f(1) = \frac{1}{2}$. And then, it follows that $f(3) = \frac{1}{3} \left(1 + \frac{1}{2}\right) = \frac{1}{2} $, etc... 
Claim: $f(n) = \frac{1}{2}$ for all $n \geq 2$. 
Proof: Assume the above holds, then $$f(n+1) = \frac{1}{n+1} \left(f(1) + \cdots + f(n) \right) = \frac{1}{n+1}\left(1 + \frac{n-1}{2}\right) = \frac{1}{n+1} \cdot \frac{n+1}{2} = \frac{1}{2}$$
and by induction, we are done. 
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