TITLE: Determining limit of recursive sequence
QUESTION [8 upvotes]: I was trying to calculate the limit of sequence defined as
$$a_1=k; a_2=l; a_{n+1}=\frac{a_n+(2n-1)a_{n-1}}{2n}; k, l\in\mathbb{N}k<l$$
I had no idea where to start with that so I've brute forced the problem on my PC for permutations of $(k, l)$ to $a_{10^{10}}$ in hope that there would be an emerging pattern. This is what my PC thinks the $a_\infty$ is for $(k, l)$:
$$(1, 2) \Rightarrow 3-\sqrt{2}$$
$$(1, 3) \Rightarrow 5-\sqrt{8}$$
$$(2, 3) \Rightarrow 4-\sqrt{2}$$
$$(1, 4) \Rightarrow 7-\sqrt{18}$$
$$(2, 4) \Rightarrow 6-\sqrt{8}$$
$$(3, 4) \Rightarrow 5-\sqrt{2}$$
$$....$$
It seems that $a_{n\rightarrow\infty}\rightarrow2l-k-(l-k)\sqrt{2}$. How can I show that mathematically, without using brute force?

REPLY [10 votes]: We have
$$a_{n+1} - a_n = \dfrac{a_n+(2n-1)a_{n-1}}{2n} - a_n = \dfrac{(2n-1)a_{n-1}-(2n-1)a_{n}}{2n} = -\dfrac{2n-1}{2n} \left(a_n-a_{n-1}\right)$$
Let $b_n = a_{n+1}-a_{n}$. We then have
$$b_{n+1} = - \dfrac{2n-1}{2n}b_n$$
with $b_1 = l-k$. We then have
$$b_{n+1}=b_1(-1)^{n} \prod_{k=1}^n \dfrac{2k-1}{2k} = b_1(-1)^n \dfrac{(2n)!}{4^n(n!)^2} = b_1\left(-\dfrac14\right)^n \dbinom{2n}n$$
We have
$$a_{n+1} -a_1= \sum_{k=1}^n \left(a_{k+1}-a_k\right) = \sum_{k=1}^nb_k = b_1 \sum_{k=0}^{n-1} \left(-\dfrac14\right)^k \dbinom{2k}k$$
Recall that
$$\sum_{k=0}^{\infty} \dbinom{2k}k x^k = \dfrac{1}{\sqrt{1-4x}}$$for $\vert x \vert \leq 1/4$. Plugging in $x=-1/4$, we obtain
$$\lim_{n \to \infty} a_{n+1} = a_1 + \dfrac{b_1}{\sqrt2}=k+ \dfrac{l-k}{\sqrt2}$$