# Finding the limit of a contractive sequence

I've been trying to solve an exercise. I am given the recursive formula of a sequence and I am asked to prove that it is converging and then find its limit.

Given $x_0=a$ and $x_1=b$, with $a,b \in R$, the recursive formula is $\displaystyle x_{n+1} = \frac{x_{n-1} +(2n-1)x_n}{2n}$.

For my solution, I showed that the difference of two successive terms is:

$|x_{n+1} - x_n| = \frac{1}{2n}|x_{n-1}-x_n|\leq\frac{1}{2}|x_{n-1}-x_n)|$ and thus the sequence is contractive. If it's contractive, it is Cauchy and if it is Cauchy then it converges.

However, I am unable to figure out how to calculate its limit. Since $a,b$ are parameters, I'm sure there is a closed formula that gives the limit and it depends on those two parameters but I can't find it. Any ideas?

Show by induction that for $n\geq 0$ $$x_{n+1}-x_n=(-1)^n\frac{b-a}{2^n n!}$$
• If you sum these equalities for $k=0,...,n$, you get $$x_{n+1}=x_0+\sum_{k=0}^n (-1)^k\frac{b-a}{2^k k!}$$ and you certainly know that $$\exp(x)=\sum_{k\geq 0} \frac{x^k}{k!}$$ Commented Dec 25, 2016 at 9:13