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Given is following sequence:

$a_{n+1} = a_n - \frac{a_n - v}{s}$

I found out that

$\forall a_0, v, s \in \mathbb{R}, s>0: \lim\limits_{n \to \infty}a_n=v$

But I do not know why. I tried to write down $a_2$ , $a_3$, but the term becomes very long and complex, and it doesn't help me to find out why the limes leads to $v$.

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  • $\begingroup$ If $\nu=0$ then the recursion becomes $a_{n+1}=(1-\frac 1s)a_n$ which is easily solved to get $a_{n}=(1-\frac 1s)^na_0$. If, say, $s=\frac 12$ then this does not have a limit. $\endgroup$ – lulu Aug 23 '15 at 13:44
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    $\begingroup$ It could be useful to compute $a_{n+1} - a_n = \frac{v - a_n}s$, which shows that if the succession converges, then it has to converge on $v$, because if it wasn't so the difference would not converge to 0, so the succession would not converge either. $\endgroup$ – Michele De Pascalis Aug 23 '15 at 13:54
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This is a first order linear recurrence relation, and the solution can be found explicitly:

$$a_n=a_0 \left(\frac{s-1}{s}\right)^n-v \left(\frac{s-1}{s}\right)^n+v.$$

Can you see that the limit doesn't necessarily exist?

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  • $\begingroup$ Thank you. This let me understand that the limit does not exist for $s\leq\frac{1}{2}$ . In all other cases, the limes is $v$ because $(\frac{s−1}{s})^n$ becomes $0$ . But can you please tell me how you did form this equation? Also, I just noticed that for $s=1$ , your equation gives $a_n = a_0$ while my original equation gives $a_n = v$ . Or is this an exceptional case? $\endgroup$ – Daniel Marschall Aug 23 '15 at 16:42
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If $s=1$ or $a_0=v$, then $a_n=v$ for all $n$ so the limit is trivial.

If $s≠1$, then let $a_n=b_n+v$. This yields: $$ b_{n+1}=b_n\left(1-\frac{1}{s}\right)\implies b_n=b_0\left(1-\frac{1}{s}\right)^n $$ So if $0.5<s$, we have $|{1-\frac{1}{s}}|<1$ and therefore $\lim_{n\to\infty}b_n=0\implies\lim_{n\to\infty}a_n=\lim_{n\to\infty}b_n+v=v$

However, if $0<s≤0.5$, $b_n$ diverges, and thus, $a_n$ diverges too.

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You can write $a_{n+1}=\left(1-\frac{1}{s}\right)a_n+\frac{v}{s}$.

The form is $a_{n+1}=Aa_n+B$, with $A\neq 1$. Now let $r=\frac{B}{1-A}$. You can show that $b_n=a_n-r$ is a geometric sequence. So $b_n=b_0A^n$ and $a_n=r+(a_0-r)A^n$. I let you substitue with the values of $A$ and $B$, so you can find the general expression of $a_n$.

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