This is a linear recursion relationship. The methods for calculating formulas for the $n$ term are well known. In particular, we can form a vector for $k$ adjacent elements of the sequence: $$U_i = \begin{bmatrix}u_{i+k}\\u_{i+k-1}\\\vdots\\u_i\end{bmatrix}$$
Then we find that $$U_{i+1} = AU_i$$ where $$A = \begin{bmatrix}\frac 1k & \frac 1k & \frac 1k & \dots & \frac 1k \\ 0 & 1 & 0 & \dots & 0 \\ 0 & 0 & 1 & \dots & 0 \\\vdots & \vdots & \vdots & \ddots & \vdots\\ 0&0&0&\dots & 1\end{bmatrix}$$
And more generally $U_{n+1} = A^nU_1$. If $A$ can be diagonalized, then there exist matrices $Q, Q^{-1}, D$ with $D$ having only diagonal entries (the eigenvalues of $A$) such that $A = QDQ^{-1}$. And therefore
$$U_{n+1} = QD^nQ^{-1}U_0$$
If $$D = \begin{bmatrix}a_k & 0 & \dots & 0 \\ 0 & a_{k-1} & \dots & 0 \\ \vdots & \vdots & \ddots & \vdots\\ 0&0&\dots & a_1\end{bmatrix}$$
which I'll abbreviate to $D = [[a_k \dots a_1 ]]$, then $D^n = [[a_k^n \dots a_1^n ]]$. So $$U_{n+1} = Q[[a_k^n \dots a_1^n ]]Q^{-1}U_1$$
Now the expression $Q[[a_k^n \dots a_1^n ]]Q^{-1}U_1$ depends linearly on each of the $a_k^n$. In particular, there exist constants $B_i$ such that $$u_n = B_1a_1^n + B_2a_2^n + \dots + B_ka_k^n$$
The characteristic polynomial of $A$ is $$x^k - \frac 1k\left(x^{k-1} + ... + x + 1\right)$$
so the eigenvalues $a_i$ are its roots. It is easily checked that the first root is $a_1 = 1$. And a look at the derivative shows that $1$ is not a multiple root. I'll leave determining the other roots to you. If they are distinct, then $A$ is diagonalizable.
Once the eigenvalues are determined, it is not necessary to figure out $Q$. Instead you can solve the system of equations formed by $$u_n = B_1a_1^n + B_2a_2^n + \dots + B_ka_k^n$$
when $n =1 \dots k$ for the constants $B_i$.
Now taking the limit $$\lim u_n = B_1\lim a_1^n + B_2\lim a_2^n + \dots + B_k\lim a_k^n$$
Provided the limits on the right-hand side converge. Since the eigenvalues may be complex, we have $4$ cases:
- $|a_i| < 1$, in which case $\lim_n a_i^n = 0$.
- $a_i = 1$, in which case $\lim_n a_i^n = 1$.
- $|a_i| = 1$ and $a_i \ne 1$, in which case $\lim_n a_i^n$ does not converge.
- $|a_i| > 1$, in which case $\lim_n a_i^n = \infty$.
For your supposition to be correct, the limit must converge to a finite value, which means you will need to show that for $i> 1, |a_i| < 1$. In this case, $$\lim u_n = B_1$$so you will also need to show that $B_1$ has the desired form.
