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Prove sequence

$$x_{n+1} = \frac{x_n+ n x_{n-1}}{n+1} $$

$$x_0 = 0, x_1 = 1 $$

converges and find it's limit

My attempt

  1. Let's prove $0 \le x_n \le 1$:

    $x_n \ge 0 $ (obvious)

    By induction

    if $x_n \le 1$ and $x_{n-1} \le 1$ then: $$ \frac{x_n+ n x_{n-1}}{n+1} \le \frac{1+ n}{n+1}=1 \implies x_{n+1}<=1$$

  2. Let's prove convergence

$$\lim_{n \to \infty }{x_{n+1}} = \lim_{n \to \infty }{ \frac{x_n + n x_{n-1}}{n+1} } = \lim_{n \to \infty }{ (\frac{x_n}{n+1} } + \frac{x_{n-1}}{1+\frac{1}{n}}) = x_{n-1}$$

So, sequence converges.

Question: I'm right so far and how to find the limit?

Thanks

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  • 2
    $\begingroup$ No, the second part is wrong. $\endgroup$ – xpaul Apr 26 '16 at 17:43
  • $\begingroup$ Yes, sequence must be decreasing or increasing on all $n$ , what's about second statement? $\endgroup$ – Evgeny Semyonov Apr 26 '16 at 17:45
  • $\begingroup$ For part 2, notice that $n$ is an "internal variable" on the left-hand side (it's used for the purpose of the finding a limit), yet it appears as a fixed value on the right-hand side (in $x_{n-1}$). This tells you that something is off. Also, you seem to assume that $\frac{x_n}{n+1} = 0$. How do you know this? (What if $x_n \approx n$, for example?) $\endgroup$ – Théophile Apr 26 '16 at 18:18
  • $\begingroup$ @Théophile Part 1: $0 \le x_n \le 1$ $\endgroup$ – Evgeny Semyonov Apr 26 '16 at 18:21
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    $\begingroup$ Ah, sorry, I missed that. In any case, my first comment still holds; if $n$ appears inside a limit on the left, it shouldn't appear outside a limit on the right. $\endgroup$ – Théophile Apr 26 '16 at 18:23
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Hint:

Check that $\displaystyle \forall n\geq 0, x_{n+1}-x_n=\frac{(-1)^n}{n+1}$

The series $\displaystyle \sum (x_{n+1}-x_n)$ is therefore convergent, and so is the sequence $(x_n)$, say $x_n\to l$

Furthermore, $\displaystyle l= \sum_{k=0}^\infty (x_{k+1}-x_k) =\sum_{k=0}^\infty \frac{(-1)^k}{k+1} = \log 2$.


Edit: from a simulation on Mathematica, here is the plot of the first few terms of the sequence, giving intuition about its behavior:

Plot for $0\leq n\leq 35$

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  • 2
    $\begingroup$ Without using series (but less immediate): Once you have the above relation between $x_{n+1}-x_n$, or even just show by induction that $x_{n+1}-x_n$ has alternating sign (e.g. by induction), it is easy to show that the sequence $(x_{2n})_n$ is increasing, and that $(x_{2n+1})_n$ is decreasing. Since both are bounded, both converge. And the limits are the same, since $x_{n+1}-x_n \to 0$. $\endgroup$ – Clement C. Apr 26 '16 at 17:54

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