# What is $\lim_{n\to\infty}a_n$ if $a_0=\alpha,a_1=\beta$ and $a_{n+1}=a_n+\frac{a_{n-1}-a_n}{2n}$?

Let $$(a_n)_{n=0}^\infty$$ be the sequence such that $$a_0 = \alpha, a_1 = \beta,$$ where $$\alpha,\beta\in\mathbb{R},$$ and $$a_{n+1} = a_n + \dfrac{a_{n-1} - a_n}{2n}$$. Find $$\lim\limits_{n\to\infty} a_n$$.

I'm not sure where to start for this problem. I got that $$a_2 = \dfrac{\alpha+\beta}{2},a_3 =\dfrac{5\beta+3\alpha}{8},$$ and $$a_4 = \dfrac{29\beta + 19\alpha}{48}.$$ I wonder if the expression for the $$n$$th term has a closed form?

Let $$b_n=a_{n+1}-a_n$$. Then $$b_n=-\frac {b_{n-1}} {2n}$$. By iteration we get $$b_n=(-1)^{n+1}\frac {b_1} {2^{n}(n!)}$$. Note that $$b_1+b_2+...+b_{n-1}=a_n-a_1$$. Hence it is enough to sum the series $$\sum b_n$$ which is an exponential series.
• $\sum b_n \approx b_1(e^{1/2}-1)$. – marty cohen Jan 1 at 2:07
• $b_n=-b_{n-1}/(2n)$. – Sungjin Kim Jan 1 at 4:18