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?

This is similar to the first answer, but without rewriting things. Note that
$$f(n+1) = \frac{1}{n+1}\sum_{x=1}^n f(x) = \frac{1}{n+1}\sum_{x=1}^{n-1}f(x) + \frac{1}{n+1}f(n).$$
Using the definition for $f(n)$, this becomes
$$f(n+1) = \frac{1}{n+1}\sum_{x=1}^{n-1} f(x) + \frac{1}{n(n+1)}\sum_{x=1}^{n-1}f(x) = \left(\frac{1}{n+1}+\frac{1}{n(n+1)}\right)\sum_{x=1}^{n-1}f(x).$$
By making a common denominator, we get that
$$\frac{1}{n+1}+\frac{1}{n(n+1)} = \frac{n}{n(n+1)}+\frac{1}{n(n+1)} = \frac{n+1}{n(n+1)} = \frac{1}{n},$$
thus
$$f(n+1) = \frac{1}{n}\sum_{x=1}^{n-1} f(x) = f(n).$$
