Limit of this recursive sequence: $x_{n+1}=\bigl(1-\frac{1}{2n}\bigr)x_{n}+\frac{1}{2n}x_{n-1}.$

Consider the following sequence :

$x_{0}=a$ , $x_{1}=b$ , $x_{n+1}=\bigl(1-\frac{1}{2n}\bigr)x_{n}+\frac{1}{2n}x_{n-1}.$ Find $\displaystyle \lim_{n\to \infty}x_{n}.$

I calculate $x_{2}$ , $x_{3}$ ,$x_{4}$ ,but I could not find any relationship between any two consecutive pair. But I found that the sum of the coefficients of $a$ & $b$ is equal to the term in denominator in each $x_{i}$.

How we find this limit...?

Solution 1:

Hint $$x_{n+1}-x_{n}=-\dfrac{1}{2n}(x_{n}-x_{n-1})$$ so $$x_{n+1}-x_{n}=(-1)^n\dfrac{1}{2^n\cdot n!}(x_{1}-x_{0})$$ so $$x_{n}-x_{0}=\sum_{i=1}^{n}(x_{i}-x_{i-1})$$ so $$\lim_{n\to\infty}x_{n}=a+\sum_{n=0}^{\infty}\dfrac{(-1)^n}{2^n\cdot n!}(b-a)=a+e^{-1/2}(b-a)$$