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

calculuslimitssequences-and-series

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?

Best Answer

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.

Best Answer

Related Solutions

Radius of convergence of $\sum_{n=0}^\infty a_n x^n$, with $a_{n+2} = \frac{n(n+1) a_{n+1} – a_n}{(n+2)(n+1)}, a_2 = -a_0/2$

Given the sequence $(a_n)_{n \ge 1}$ with $a_1=2$ and $a_{n+1} = \frac{n^2-1}{a_n} + 2$ for $n \ge 1$, find the following limits.

Related Question