let $x_0 = a$ and $x_1 = b $ , $x_{n+1} = ( 1- \frac{1}{2n}) x_n + \frac{1}{2n} x_{n-1}$, $n\ge 1$ find $lim_{n \rightarrow \infty}x_n$?

My attempt: I take $x_n = a + (b-a) +\dots+ (x_n - x_{n-1})$

after that that I am not able to proceed further.

Please help me or any hints/solution will be appreciated.

Thank you and thanks in advance for giving hints/solutions

  • 1
    $\begingroup$ Hint: find the formula for $x_{n+1} - x_n$, and see if their sum looks familiar. $\endgroup$ – user58697 Jun 7 '18 at 20:39
  • 4
    $\begingroup$ A hint: Observe that $$x_{n+1}-x_n=-\frac{1}{2n}(x_n-x_{n-1}).$$ $\endgroup$ – user539887 Jun 7 '18 at 20:41

Recall that a sequence in $\mathbb{R}$ converges if and only if it is a Cauchy sequence. What's more the recursive definition gives:

$\vert x_{n+1}-x_n\vert=\frac{1}{2n}\vert x_n-x_{n-1}\vert=...=\frac{1}{2^n \cdot n!} \vert b-a\vert $

Which then means that for all $n,k\in \mathbb{N}$:

$\vert x_{n+k}-x_n \vert \leq \overset{k}{\underset{l=1}{\sum}}\vert x_{n+l}-x_{n+l-1}\vert\leq \vert b-a\vert \cdot \overset{k}{\underset{l=1}{\sum}}\frac{1}{2^{n+l} \cdot (n+l)!}\leq \vert b-a\vert \cdot \overset{k}{\underset{l=1}{\sum}}\frac{1}{2^{n+l}}\ $

And for all $\epsilon>0$ and $n$ large enough we'll obtain that:

$\vert x_{n+k}-x_n \vert <\epsilon$

Which means that the sequence is Cauchy.


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