The limit of averages

In dependence on the first two members \( a_0, a_1 \) of the sequence, determine the limit \(\lim_{n\to\infty} a_n \).

Variant
In the sequence \( (a_0, a_1, a_2, \ldots) \) every further element is always the arithmetic mean of the previous two elements (i.e. \(a_{n+2}=\frac{a_{n+1}+a_n}{2}\) for \(n\geq 0 \)).
Solution
The charakteristic polynomial is \(\lambda^2-\frac12 \lambda-\frac12=0\) with roots \(\lambda_1=1\), \(\lambda_2=-\frac12\).

The analytic expression is \(a_n=\alpha\left(-\frac12\right)^n+\beta\).

From the equations for \(a_0\) and \(a_1\) we get \(\beta=\frac{a_0+2a_1}{3}\) and \(\alpha=\frac{2a_0-2a_1}{3}\).
Answer
The limit is \(\frac{a_0+2a_1}{3}\).
Variant
In the sequence \( (a_0, a_1, a_2, \ldots) \) every further element is always the arithmetic mean of all the previous elements (i.e. \(a_n=\frac{a_0+\ldots+a_{n-1}}{n}\) for \(n\geq 0 \)).
Answer
The limit is \(\frac{a_0+a_1}{2}\), because the sequence is constant from \(a_2\).