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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?

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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.

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  • $\begingroup$ $\sum b_n \approx b_1(e^{1/2}-1)$. $\endgroup$ – marty cohen Jan 1 at 2:07
  • $\begingroup$ $b_n=-b_{n-1}/(2n)$. $\endgroup$ – Sungjin Kim Jan 1 at 4:18
  • $\begingroup$ @SungjinKim Thanks for the comment. $\endgroup$ – Kavi Rama Murthy Jan 1 at 5:16

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