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Give an example of a sequence which is neither increasing after a while, nor decreasing after a while, yet which converges to 1.

My solution: $1.01,\ .99,\ 1.001,\ .999,\ 1.0001,\ .9999,\ \text{etc}\dots$

Does that satisfy all the conditions? Also, judging by the instructions, do you think I would have to define that sequence? In which case, I could do $\{x_n\} = 1 + .01^n$ for odd $n$ and $1 - .01^n$ for even $n$ (which would change the sequence, but just increases the rate at which it approaches $1$).

The definition of an increasing sequence used is the next term being bigger than OR equal to the preceding term. And dually for decreasing.

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    $\begingroup$ $1.1,\ 1.2,\ 1.01,\ 1.02,\ 1.001,\ 1.002,\ 1.001,\ 1.002,\ 1.0001,\ 1.0002,\dots$ $\endgroup$
    – bof
    Commented Sep 20, 2016 at 12:11
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    $\begingroup$ The term you are looking for is "non-monotonic at the limit" $\endgroup$
    – BlueRaja - Danny Pflughoeft
    Commented Sep 20, 2016 at 16:32
  • $\begingroup$ Limits superior and inferior are nice for these things, to split them into 2 different sequences. $\endgroup$
    – Simply Beautiful Art
    Commented Sep 20, 2016 at 21:57
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    $\begingroup$ Yes, you got it. But I think your series is $1 + (-0.1)^n$. $\endgroup$
    – Steven Alexis Gregory
    Commented Sep 21, 2016 at 5:55

7 Answers 7

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I think defining it the way you did is fine, but you could also use

$$s_n = 1+\left(-\frac{1}{10}\right)^n$$

which is a handy little trick to express alternating sequences nicely.

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    $\begingroup$ I think it just has to be a sequence that doesn't increase or decrease the whole time (1.1, 1.01, 1.001 or .9,.99,.999) $\endgroup$
    – Remy
    Commented Sep 20, 2016 at 2:58
  • $\begingroup$ But I could be reading the instructions incorrectly. $\endgroup$
    – Remy
    Commented Sep 20, 2016 at 2:59
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    $\begingroup$ Oh, I just realized. 1,1,1,1,1 is both increasing and decreasing. Since the definition given by my prof for increasing is the next term being bigger than or equal to the preceding term. $\endgroup$
    – Remy
    Commented Sep 20, 2016 at 3:32
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    $\begingroup$ @Brick : The instruction "neither increasing ..., nor decreasing ..." means that there is no finite initial segment after which the sequence only decreases or only increases. No tail of this answer's sequence only decreases or only increases. $\endgroup$
    – Eric Towers
    Commented Sep 20, 2016 at 5:30
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    $\begingroup$ You can make the expression lower with another minus in the exponent: $$1+(-10)^{-n}$$ $\endgroup$
    – CiaPan
    Commented Sep 20, 2016 at 10:38
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Your solution works.

I like $1+\dfrac{\sin n}{n}$. But it's possible I'm weird.

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  • $\begingroup$ I was going to post this too, but someone other than op edited in criteria for an increasing/decreasing function, so depending on how strict the initial requirements are, this may or may not work. $\endgroup$
    – Burnsba
    Commented Sep 20, 2016 at 12:17
  • $\begingroup$ @Burnsba : Since no two adjacent terms of this sequence are equal, the non-strictness of "increasing" and "decreasing" does not impact its validity. (We would look for $k \in \Bbb{Z}$ such that $\dfrac{(2k+1)\pi}{2} \pm \dfrac{1}{2}$ are both integers, but this would require that the transcendental number $\dfrac{(2k+1)\pi}{2}$ is rational, which is absurd.) $\endgroup$
    – Eric Towers
    Commented Sep 20, 2016 at 12:51
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    $\begingroup$ @Burnsba : The instruction "neither increasing ..., nor decreasing ..." means that there is no finite initial segment after which the sequence only decreases or only increases. No tail of this answer's sequence only decreases or only increases. If we were to take your interpretation, no solution is possible. $\endgroup$
    – Eric Towers
    Commented Sep 20, 2016 at 13:05
  • $\begingroup$ This is what I was going to suggest. $\endgroup$
    – Brian
    Commented Sep 20, 2016 at 20:11
  • $\begingroup$ I like the intuitive appeal, but it is admittedly kind of subtle to get the exact proof for this one -- a bit of a rabbit hole dive $\endgroup$
    – MichaelChirico
    Commented Sep 21, 2016 at 2:29
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EDIT: As the comments note, the original formulation of the question was not specific about whether the inequalities in the definitions were strict. With the additional specification now that "increasing" and "decreasing" allow for equality, this answer of course doesn't fit anymore.

You could just take the sequence 1, 1, 1, 1, 1, ... .

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    $\begingroup$ Oh, true... I didn't think of that. I guess that would work since the definition of an increasing sequence is the next term being bigger than OR equal to the preceding term. $\endgroup$
    – Remy
    Commented Sep 19, 2016 at 23:41
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    $\begingroup$ That technically works, although the question is rather vague about strict vs. non-strict increasing and decreasing. $\endgroup$
    – Carl Schildkraut
    Commented Sep 19, 2016 at 23:41
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    $\begingroup$ I’m sure some people do take “strictly in/decreasing” as their default reading of “in/decreasing sequence”, in which context this answer is very nice. But it’s much more common in my experience for “in/decreasing sequence” to mean the notion with ≤ or ≥, in which case this answer doesn’t satisfy it. $\endgroup$
    – Peter LeFanu Lumsdaine
    Commented Sep 20, 2016 at 7:41
  • $\begingroup$ @RemyM " I guess that would work since the definition of an increasing sequence is the next term being bigger than OR equal to the preceding term." But this sequence has the property that the next term is "bigger than OR equal to the preceding term." Thus it is an increasing sequence and not an example of what you are looking for. $\endgroup$
    – quid
    Commented Sep 20, 2016 at 11:46
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    $\begingroup$ @quid I've always seen increasing to mean strictly increasing, but if that's not the usual meaning then I guess this answer wouldn't work. $\endgroup$
    – Carl Schildkraut
    Commented Sep 20, 2016 at 13:59
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I think it is possible to use a well known Fibonacci sequence property:

$$\lim_{n\to\infty}\frac{F_{n}}{F_{n-1}} = \varphi$$

where $\varphi$ is the golden ratio, and also it is well known that each term of the sequence, i.e. let us call it $a_n=\frac{F_{n}}{F_{n-1}}$, oscillates between a value over and under $\varphi$. For that reason the sequence $$\left\{n \gt 2:a_n=\frac{F_{n}}{F_{n-1}\cdot \varphi}\right\}$$ will be neither increasing after a while, nor decreasing after a while, and will converge to $1$.

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    $\begingroup$ You can turn this into a rational sequence by dividing two consecutive terms: $b_n = F_{n-1} F_{n+1} / F_n^2$. $\endgroup$
    – Erick Wong
    Commented Sep 20, 2016 at 10:44
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    $\begingroup$ I've made a little fix into your post. If you prefix lim with a backslash you convert it into a LaTeX symbol, which is properly rendered in upright font: \lim → $\lim$, and has a limit condition properly put underneath it, instead of italics, which looks like a $l\cdot i\cdot m$ multiplication: lim → $lim$ with a condition appended to 'em' . $\endgroup$
    – CiaPan
    Commented Sep 20, 2016 at 10:46
  • $\begingroup$ @CiaPan thank you for that change! $\endgroup$
    – iadvd
    Commented Sep 20, 2016 at 11:05
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Yes, your sequence satisfies all required conditions.

Another example sequence: $$a_n = \begin{cases} 1+\frac 1n & \text{if } \log_{10}n \in \mathbb N \cup \{0\} \\ 1 & \text{otherwise} \end{cases}$$ has terms: $$\begin{align} a_1 & = 2 \\ a_{10} & = 1.1 \\ a_{100} & = 1.01 \\ a_{1000} & = 1.001 \\ a_{10000} & = 1.0001 \\ \ldots \end{align}$$ and all remaining terms equal $1$.

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It's so easy to find such a series, for example: $$a_n=q^n+1\ ,\ -1<q<0$$

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    $\begingroup$ Easy. Once you know how. Give a discoverer a chance to discover. $\endgroup$
    – fleablood
    Commented Sep 20, 2016 at 0:06
  • $\begingroup$ You're right, I just wanted to state that basically it's not a difficult question, but an easy one. $\endgroup$
    – 76david76
    Commented Sep 20, 2016 at 4:34
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    $\begingroup$ It is easy. But I'm impressed the op's reasoning is spot on, albeit niave. $\endgroup$
    – fleablood
    Commented Sep 20, 2016 at 4:58
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Continued fraction representations of irrational numbers can be represented as an infinite sequence of rational partial convergents which oscillate above and below the number being represented.

Thus if we take $a_n$ to be the to be the nth partial convergent of any irrational number divided by that number, it will converge to 1 in an oscillatory manner.

$$ a_n = \frac{[\zeta]_n}{\zeta}, \zeta \in \mathbb{I} $$

For $\zeta = \sqrt{2}$, we have

$$ a_0 = \frac{1}{\sqrt{2}} \approx 0.707\\ a_1 = \frac{3}{2\sqrt{2}} \approx 1.061\\ a_2 = \frac{7}{5\sqrt{2}} \approx 0.990\\ a_3 = \frac{17}{12\sqrt{2}} \approx 1.002\\ ... $$

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