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I have trouble understanding why, when finding the limit of a recurrently defined sequence, we can assume that $x_n=x_{n+1}$ to find the actual limit.

I figured it's got something to do with the fact that $\lim\limits_{n \to \infty} x_n=\lim\limits_{n \to \infty} x_{n+1}$ and intuitively that makes perfect sense, but it's the theory behind it that evades my understanding (we can't just say that "because $|x_n-x_{n+1}|$ for high enough $n$ is very close to zero, we might aswell make them equal", right?)

Example: In $x_1=0$ , $ x_{n+1}=\frac{1}{1+x_n}$ I'd use the equality $ L=\frac{1}{1+L}$

This question probably isn't very well formulated, but that comes from my lack of understanding of the problem, sorry! (and thanks for the answer :)

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We don't really assume that $x_n=x_{n+1}$. Instead, suppose that the limit exists, call it $x$. Then $\lim_{n\to\infty}x_n=x$. This, by definition of limit, means that for every $\epsilon>0$ there is a $N\in\mathbb N$ such that for all $n>N$ we have $|x_n-x|<\epsilon$. Intuitively, for every $\epsilon>0$, the sequence is $\epsilon$-close to $x$ from some place on.

Now, we will show that it follows that $\lim_{n\to\infty}x_{n+1}=x$. Let $\epsilon>0$. Then for $n>N$ we have $n+1>N$ and therefore by the previous paragraph, $|x_{n+1}-x|<\epsilon$. So, for all $n>N$, the inequality $|x_{n+1}-x|<\epsilon$ holds. This means that for every $\epsilon>0$, the number $x_{n+1}$ is also $\epsilon$-close to $x$ for all $n>N$, i.e. the limit exists and is equal to $x$.

Similarly, we can show that if $\lim_{n\to\infty}x_{n+1}=x$, then $\lim_{n\to\infty}x_n=x$. To see this, just write out what $\lim_{n\to\infty}x_{n+1}=x$ means in terms of $\epsilon$, $N$ and $n$. Then you will be able to show that $x_n$ is $\epsilon$-close to $x$ for all $n>N+1$.

What all this means is that $\lim_{n\to\infty}x_n=x$ holds precisely if $\lim_{n\to\infty}x_{n+1}=x$ holds, i.e. the conditions are equivalent.

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    $\begingroup$ Perfect, thanks a lot! $\endgroup$
    – Dahn
    Nov 11, 2012 at 13:09
  • $\begingroup$ So, if $\lim_{n \to \infty}(u_{n} -u_{n+2}) = 0$, then $\lim_{n \to \infty}(u_{n} -u_{n+1}) = 0$? $\endgroup$
    – Mr.Lilly
    Jul 11, 2016 at 14:32
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    $\begingroup$ @nameless: that's true, if you assume for instance that $\lim_{n\to\infty}u_n$ exists. Otherwise, $u_n=(-1)^n$ is a counterexample. $\endgroup$
    – Dejan Govc
    Jul 11, 2016 at 16:03
  • $\begingroup$ @DejanGovc Thanks for your reply. Please check for the proof. We assume that $\lim_{n} u_{n}$ exists, namely, $L \in \mathbb{R}$. Let $\epsilon > 0$. By our assumption, there exists natural number $K$ such that for each $n \geq K$, we have $|u_{n} - L| < \frac{\epsilon}{2}$. Then $|u_{n} - u_{n+1}| \leq |u_{n} - L| + |u_{n+1} - L| < \epsilon$. Is this correct? If it is correct, when we use the condition $\lim_{n \to \infty}(u_{n} -u_{n+2}) = 0$? $\endgroup$
    – Mr.Lilly
    Jul 15, 2016 at 17:11
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    $\begingroup$ @nameless: looks good to me. You don't have to use $\lim_{n \to \infty}(u_{n} -u_{n+2}) = 0$. If the limit exists, $\lim_{n \to \infty}(u_{n} -u_{n+1}) = 0$ and $\lim_{n \to \infty}(u_{n} -u_{n+2}) = 0$ are automatically true. (In particular, they imply each other.) $\endgroup$
    – Dejan Govc
    Jul 15, 2016 at 17:24

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