Given $a,b\in\mathbb R$ , defined $(a_n)_n$ recursively by setting
$a_1 = a$ ,$ a_2 = b$ ,$ a_{n+1} = \frac{1}{2n}a_{n-1}+\frac{2n-1}{2n}a_n,\;\;n\geq 2 $
Determine $\displaystyle\lim_{n \to \infty} a_n$
My attempt : I know that $a_n = a + (b-a) +.......+ (a_n - a_{n-1})$ after that I'm not able to procede further.
any hints/solution will be appreciated
Thank you!
Hint: $$ a_{n+1} - a_n = - \frac{1}{2n}(a_n - a_{n-1}) $$
Thus $a_{n+1}-a_n = \frac{(-1)^{n-1}}{(2n)!!}(b-a)$
$$ a_n - a = \sum_{i=1}^{n-1}(a_{i+1}- a_{i}) = (a - b) \sum_{i=1}^{n-1} \frac{(-1)^i}{(2i)!!} $$
The summation certainly converges, but I don't know how to get the converged value
According to wolframalpha, (and thanks to @Minus One Twelfth hint) $$ \sum_{i=1}^{\infty} \frac{(-1)^i}{(2i)!!} = \sum_{i=0}^{\infty} \frac{1}{i!}\left(-\frac{1}{2}\right)^i - 1 = e^{-1/2} - 1 $$
In summary, $\lim_{n \rightarrow \infty} a_n = \left(1-\frac{1}{\sqrt{e}}\right) b + \frac{1}{\sqrt{e}}a$