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$x_0=a$, $x_1=b$ define $$x_{n+1}=\left(1-\frac{1}{2n}\right)x_n+ \frac{1}{2n} x_{n-1}$$

we need to find the limit of $\{x_n\}_n$ as $n\rightarrow \infty$.

Let $l$ be this limit , I tried taking the limit on both sides getting $l=l$ as $n\rightarrow \infty$.

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The reason is that the limit seems, from a few numerical experiments, to depend on $a$ and $b$. Actually you don't get any necessary condition about $l$, it may be anything and you should also prove that it exists. – Siminore Oct 3 '12 at 9:16
up vote 8 down vote accepted

Let $d_n =x_{n+1}-x_n$. Then $d_0=b-a$ and from the recursion formula $$d_n=x_{n+1}-x_x=\frac1{2n}(x_{n-1}-x_n) = -\frac1{2n}d_{n-1}.$$ We conclude that $$ d_n = \left(-\frac12\right)^n\frac1{n!}(b-a)$$ and we thus find $$\lim_{n\to\infty}x_n = x_0 + \sum_{k=0}^\infty d_k=a+(b-a)\sum_{k=0}^\infty \frac1{k!}\left(-\frac12\right)^k=a+\frac{b-a}{\sqrt e}.$$

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HINT: Try $y_{n+1}=x_{n+1}-x_n$ so that $y_{n+1}=-\cfrac {y_n}{2n}$

Then sum the $y_n$ (which are easy to identify) and work with the series you get.

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As $x_{n+1}=(1-\frac{1}{2n})x_{n}+\frac{1}{2n}x_{n-1}$, we can get that:

$$\begin{align*} x_{n+1}-x_{n}&=-\frac{1}{2n}(x_{n}-x_{n-1})\\ x_{n}-x_{n-1}&=-\frac{1}{2(n-1)}(x_{n-1}-x_{n-2})\\ &\vdots\\ x_{2}-x_{1}&=-\frac{1}{2}(x_{1}-x_{0})\\ \end{align*}$$

then,by computation $$x_{n+1}-x_{n}=\frac{(-\frac{1}{2})^{n}}{n!}(b-a)$$

$$\begin{align*} x_{n}&=\sum_{k=0}^{n-1}(x_{k+1}-x_{k})+x_{0}\\ &=a+(b-a)\sum_{k=0}^{n-1}\frac{(-\frac{1}{2})^{k}}{k!}\\ \end{align*}$$

$$\begin{align*} \lim_{n\rightarrow\infty}x_{n} &=a+(b-a)\sum_{k=0}^{\infty}\frac{(-\frac{1}{2})^{k}}{k!}\\ &=a+(b-a)e^{-\frac{1}{2}}\\ &=(1-e^{-\frac{1}{2}})a+e^{-\frac{1}{2}}b \end{align*}$$

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We have $$ y_{k+1}:=x_{k+1}-x_k=-\frac{x_k-x_{k-1}}{2k}=-\frac{y_k}{2k} \quad \forall k \ge 1. $$ It follows that $$ y_{n+1}\prod_{k=2}^n y_k=\prod_{k=1}^n y_{k+1}=\frac{(-1)^n}{n!2^n}\prod_{k=1}^n y_k, $$ i.e. $$ x_{n+1}-x_n=y_{n+1}=\frac{(-1)^n y_1}{n!2^n}=\frac{(-1)^n}{n!2^n}(b-a) \quad \forall n \ge 0. $$ Thus $$ x_n-x_0=\sum_{k=0}^{n-1}(x_{k+1}-x_k)=(b-a)\sum_{k=0}^{n-1}\frac{1}{k!}\cdot\left(-\frac{1}{2}\right)^k. $$ Hence $$ \lim_nx_n=x_0+(b-a)\sum_{k=0}^\infty\frac{1}{k!}\cdot\left(-\frac{1}{2}\right)^k =a+(b-a)e^{-1/2}. $$

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