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

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 $1\over\sqrt2$.

  • $\begingroup$ 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

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$\,a_{n+1} \ge a_n\,$

  • $\begingroup$ 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

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 $\displaystyle a_n\to\frac1{\sqrt{2}}$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.