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Define

$$ a_{n+1} = {1\over2\bar a_n} $$

where $\bar a_n$ is the average of $a_1, \dots, a_n$ and $a_1 = 1$. It is easy to see that $\lim a_n = {1\over\sqrt2}$, provided the limit exists. It does, indeed, as the sequence is increasing for $n>1$, except that I couldn't prove this (apparently simple) fact. Note also that the averages decrease approaching the limit from the right.

I've proven that the sequence converges based on:

  1. The sequence is bounded (use induction to see that ${1\over2}\le a_n\le1$).
  2. If a subsequence $a_{n_k}$ has a limit $\ell$, then the subsequence of averages $\bar a_{n_k} = (a_1 + \cdots + a_{n_k})/n_k$ also converges, and it does to $1\over2\ell$.
  3. Idem 2, but removing all the $a_{n_i}$ from the sum and taking their average (assume $k/n_k \to 0$ here).
  4. Any other convergent subsequence $a_{n'_k}$ must converge to the same $\ell$ (remove both the $a_{n_i}$ and the $a_{n'_i}$ from the sum).
  5. It follows that the sequence can only have one accumulation point.

The part I've been unable to prove is that $a_n$ increases for $n>1$.


Addendum: Origin of the problem

Here I will describe the origin of the $a_n$ sequence.

In the Oil Industry sometimes the rate at which a well produces depends on the rate at which water is injected back into the reservoir, a technique known as voidage replacement.

It may happen, however, that the ideal injection rate cannot be satisfied. In consequence, the expected production rate is reduced by a factor $\varphi$, which is the quotient between the actual voidage replacement fraction (total injection/total production) to the required voidage replacement fraction.

In a discrete simulation of this technique, assume that the actual injection rate is $1\over2$ of the injection requirement $r$. This means that at every discrete simulation step (e.g., 1 day) we have to penalize the ideal production rate $q$ according to the quotient defined above. This is what happens:

  1. We start with $\varphi=1$ and produce $q$ the first day. However, we inject $r\over2$ instead of $r$.
  2. The production penalty for the second step is $$ \varphi = {{r/2\over q}\over{r\over q}} = {1\over2} $$ and therefore we produce ${1\over2}q$.
  3. In the third step the factor is $$ \varphi = {{r/2 + r/2\over q+q/2}\over{r\over q}} = {1\over2(1 + {1\over2})} $$

and so on.

The resulting sequence of values of $\varphi$ behaves exactly as our definition of $(a_n)_n$. In the long run, the production is penalized by a factor that quickly and increasingly approaches $1\over\sqrt2$.

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  • $\begingroup$ The average will be increasing only if the numbers $a_1, a_2, \dots, a_n$ are positive. Because, let us consider the average of $1, 2, 3$ and $1, 2, 3, -4$. Then, the average of $1, 2, 3$ is $2$ and the average of $1, 2, 3, -4$ is $\dfrac{1}{2}$ which does not follow the increasing trend. The increasing trend would be followed (and hence, your statements would be correct) if these numbers considered in the sequence are positive. Note: I have not tried. This is just an intuitive idea! $\endgroup$ – Aniruddha Deshmukh Jan 4 '18 at 16:59
  • $\begingroup$ @AniruddhaDeshmukh I think it is positive since $$\begin{align} a_1 &= 1 \\ a_2 &= \frac{1}{2a_1} = \frac{1}{2} \\ a_3 &= \frac{1}{2\frac{a_1 + a_2}{2}} = \frac{2}{3} \\ \vdots \end{align}$$It can be shown by induction that all terms are positive. $\endgroup$ – Alex Vong Jan 4 '18 at 17:16
  • $\begingroup$ @AlexVong: See the hint I've added to item 1. $\endgroup$ – Leandro Caniglia Jan 4 '18 at 20:08
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Start with $\;\displaystyle \frac{n}{2a_{n+1}} = a_1+\ldots+a_{n-1}+a_n = (a_1+\ldots+a_{n-1})+a_n = \frac{n-1}{2a_{n}}+a_n\,$, then:

  • $\;\displaystyle \frac{1}{a_{n+1}} = \frac{n-1}{na_{n}}+\frac{2a_n}{n} \ge \sqrt{2}\,$ by induction for $\,n \ge 2\,$
    because $\,2a_n^2-\sqrt{2}n a_n + n-1$ $\displaystyle=2 \left(a_n - \frac{1}{\sqrt{2}}\right)\left(a_n - \frac{n-1}{\sqrt{2}}\right)$ $\ge 0\,$ when $\displaystyle\,a_n \le \frac{1}{\sqrt{2}}\,$

  • $\;\displaystyle \frac{1}{a_{n+1}} - \frac{1}{a_{n}} = -\frac{1}{na_{n}}+\frac{2a_n}{n} \le 0\,$ since $\displaystyle\,a_n^2 \le \frac{1}{2}\,$, and therefore$\,a_{n+1} \ge a_n\,$

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  • $\begingroup$ You've omitted the bars? i.e. $\bar a_{n+1}$. $\endgroup$ – Myridium Jan 4 '18 at 21:28
  • $\begingroup$ @Myridium There are no bars in the above. What is being used is the definition of those averages $\,\overline{a_n}= \frac{1}{n}(a_1+a_2+\ldots+a_n)\,$. $\endgroup$ – dxiv Jan 4 '18 at 21:31
  • $\begingroup$ Alright, I see. $\endgroup$ – Myridium Jan 4 '18 at 21:32
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You asked for a proof of monotony, but you mentioned other attempts to prove the convergence of your sequence, so I think it might be interesting to give one not using monotony.
With $s_n=a_1+\ldots+a_n$, we have $$s_{n+1}=s_n+\frac{n}{2\,s_n},$$ squaring both sides gives $$s^2_{n+1}=s^2_n+n+\frac{n^2}{4\,s^2_n}.\tag1$$ This implies $s^2_{n+1}>s^2_n+n$, i.e. $$s^2_n\ge s_1^2+\sum^{n-1}_{k=1}k=s_1^2+\frac{n(n-1)}2.\tag2$$ Plugging this into (1) gives $\displaystyle s^2_{n+1}<s^2_n+n+\frac{n}{2\,(n-1)}=s^2_n+n+\frac12+\frac1{2\,(n-1)}$, meaning $$s^2_n\le s^2_1+\frac{n(n-1)}2+\frac{n-1}2+\frac12\,H_{n-2}=s^2_1+\frac{n^2-1}2+\frac12\,H_{n-2},\tag3$$where $H_n$ are the harmonic numbers. (2) and (3) together with the squeeze theorem give us $\displaystyle\frac{s^2_n}{n^2}=\bar{a_n}^2\to\frac12$ as $n\to\infty$, and so we have $\displaystyle a_n\to\frac1{\sqrt{2}}$.

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  • $\begingroup$ Very smart! Thanks a lot. Note also that the steps 1 to 5 in my question draft a proof, clearly different than yours (which I liked more). $\endgroup$ – Leandro Caniglia Jan 5 '18 at 13:15

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