Show that the limit of $u_{n+d} = \frac{u_n + \cdots + u_{n+d-1}}{d}$ is $\frac{2}{d(d+1)}(u_0 + 2u_1 +\dots + du_{d-1})$

Question

Let $u_0,\ldots,u_{d-1} \in \Bbb R$.

We define $(u_n)$ by the recursive relation: $$ u_{n+d} = \frac{u_n + \cdots + u_{n+d-1}}{d} $$ Show that $$\lim_{n\to +\infty} u_n = \frac{2}{d(d+1)}(u_0 + 2u_1 + \cdots + du_{d-1})$$

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I solved it for $d=2$ by looking at $u_{n+2}-u_{n+1}$ but I think the generalization is trickier. I don't really know where to start.

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Edit:

I read the solution proposed by the first comment. But I'm looking for another type of solution. I found this question in an exam for students that ask as a preliminary question to prove Gauss-Lucas theorem. I really wonder where is the link between both questions?

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Edit 2: Edit 3:

Since:

$$ |u_{n+d}| \leq \frac{|u_n| + \cdots + |u_{n+d-1}|}{d} \leq \max\{|u_n|;\cdots |u_{n+d-1}|\} $$

it is easy to show by recurrence that $\forall n, |u_n|\leq \max \{|u_0|;\cdots |u_{d-1}|\}$.

Thus, $u_n$ is bounded.

However, if $P$ denotes the characteristic polynomial, $d \times P = d X^d - X^{d-1} - ... - 1$. And $P(z) = 0 \implies |z|^d \leq (1/d) (|z|^0 + ... + |z|^{d-1}$. Thus, we easily get that $|z|\leq 1$ and $|z| = 1 \iff z = 1$ (since $-1$ cannot be a root).

Thus, if $u_n = \sum_{\lambda ; P(\lambda)=0} \alpha_{\lambda,n}\lambda^n$ is the solution of the reccurence, as $n\to \infty$, $u_n \sim \alpha_{1,n}$. But $\alpha_{1,n}$ is polynomial in $n$ and $u_n$ is bounded. Thus $\alpha_{1,n}$ is a constant and is the limit.

Answer 1

(Inspired from here) Let me use backward indexes. We have

$$ d \, u_{n} = \sum_{j=1}^{d} u_{n-j}$$

Summing $\sum_{k=1}^{d-1} (d-k) u_{n-k}$ on each side: $$\begin{align} \sum_{k=0}^{d-1} (d-k) u_{n-k} &= \sum_{j=1}^{d} u_{n-j} + \sum_{k=1}^{d-1} (d-k) u_{n-k}\\ &=\sum_{j=1}^{d} (d-j+1) u_{n-j}\\ &= \sum_{k=0}^{d-1} (d-k) u_{n-k-1} \end{align} $$

Then, calling $y_n=\sum_{k=0}^{d-1} (d-k) u_{n-k}$ we have $y_n=y_{n-1}$, hence $y_n$ is a constant.

In particular, $y_\infty = y_{d-1} = u_0 + 2u_1 + \cdots + du_{d-1}$

From this we get the limit value of $u_n$:

$$ y_\infty= u_\infty \frac{d(d+1)}{2 } \implies u_\infty = \frac{2}{d(d+1)}(u_0 + 2u_1 +\dots + du_{d-1})$$

(Caveat: this assumes the existence of the limit).