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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!

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  • $\begingroup$ May I ask where you found that task? (: $\endgroup$
    – Invisible
    Dec 28 '19 at 17:56
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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, $\lim_{n \rightarrow \infty} a_n = \left(1-\frac{1}{\sqrt{e}}\right) b + \frac{1}{\sqrt{e}}a$

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  • $\begingroup$ Why from this we'll get a convergence? $\endgroup$ Mar 1 '19 at 7:37
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    $\begingroup$ By the way, for the infinite sum, we can use the fact that $\color{blue}{(2i)!! = 2^{i}i!}$ and then use the Taylor series for $e^{x}$. $\endgroup$ Mar 1 '19 at 8:10
  • $\begingroup$ Thanks @MinusOne-Twelfth, that solves the summation analytically $\endgroup$
    – MoonKnight
    Mar 1 '19 at 8:21
  • $\begingroup$ @MoonKnight Can we prove $a_{n+1}-a_n = \frac{(-1)^{n-1}}{(2n)!!}(b-a)$ by descending induction? That way I got $\frac{1}{2n\cdot2(n-1)\cdot2(n-2)\cdot\ldots}$ when adding all the terms which is the pattern for the Taylor series. $\endgroup$
    – Invisible
    Dec 28 '19 at 18:56

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