let us moving to telescopic sum using exponent ,Assume we have this sequence: $a_n=(1-\frac12)^{(\frac12-\frac13)^{...^{(\frac{1}{n}-\frac{1}{n+1})}}}$ with $n\geq1$ , this sequence can be written as power of sequences : ${x_n} ^ {{{y_n}^{c_n}}^\cdots} $ such that all them value are in $(0,1)$, I want to know if the titled sequence should converge to $1$ ? and how we can evaluate it for $n$ go to $\infty$ ?

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    $\begingroup$ I doubt it goes to $1$. If it did, then presumably as similar argument would show that $$b_n=(\frac12-\frac13)^{(\frac13-\frac14)^{...^{(\frac{1}{n}-\frac{1}{n+1})}}}\to1,$$ but then $a_n\to (1-\frac12)^1=1/2$. $\endgroup$ – saulspatz Jun 16 '18 at 23:12
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    $\begingroup$ @zeraouliarafik Wowie, what a cool problem. May I ask how this came up? Was it just your own curiosity, or did a friend post it to you? $\endgroup$ – Franklin Pezzuti Dyer Jun 16 '18 at 23:13
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    $\begingroup$ No, it is my own problem , The motivation is to look what about telescopic using power $\endgroup$ – zeraoulia rafik Jun 16 '18 at 23:14
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    $\begingroup$ @user625 Nope, your recursion is wrong. Your recursion would give the sequence $$a_n=(1/2)^{(1/6)(1/12)(1/20)...(1/n(n+1))}$$ $\endgroup$ – Franklin Pezzuti Dyer Jun 16 '18 at 23:24
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    $\begingroup$ @Frpzzd you are right. $\endgroup$ – user625 Jun 16 '18 at 23:41

Numeric calculation of the sequence $\{a_n\}_{n \ge 1}$ suggests that the terms are bounded, but alternate between approximately $$0.56778606544394002098000796382530333102219963214866$$ and $$0.85885772008416606762434379473241623070938618180813,$$ but I do not have a proof. This convergence is extremely rapid, and the alternating nature suggests that it is important to look at even and odd $n$ separately.

  • $\begingroup$ Would you mind also posting two plots of $a_n$, one for odd and one for even $n$? $\endgroup$ – Szeto Jun 19 '18 at 8:26
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    $\begingroup$ This dichotomy is clear from just looking at the sequence since the first power (using up the last two exponents) is $\epsilon^{\epsilon'}\simeq 1$, then the next one is $\simeq\epsilon''^1$ and so on in an alternating pattern, until one arrives at roughly $(1/2)^1$ or $(1/2)^{1/6}$. $\endgroup$ – user138530 Jun 20 '18 at 3:47
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    $\begingroup$ Also it is clear (by induction) that the "odds" go up and the "evens" go down, so the only problem (unless you insist on exact values of the limits, which is most likely out of question) is to show that they stay separated. $\endgroup$ – fedja Jun 21 '18 at 13:05
  • $\begingroup$ That sequence is already assigned A328941 and A328942 $\endgroup$ – zeraoulia rafik Nov 1 '19 at 16:03

This only shows that the limit cannot be $1$.

Note that $a_n=(1/2)^{(1/6)^{(1/12)^\cdots}}$, where the "$\cdots$" are meant to terminate at the exponent $1/(n(n+1))$.

As a general rule, if $0\lt r\lt1$ and $0\lt a\lt b\lt1$, then $0\lt r\lt r^b\lt r^a\lt1$. It follows that


and thus also that


so that, finally,


These bounds accord with what heropup found.

  • $\begingroup$ Did you miss $3\cdot 4 =12$ ? $\endgroup$ – Diger Jun 21 '18 at 1:38
  • $\begingroup$ @Diger, yes, you're right. I'll fix it. Thank you! $\endgroup$ – Barry Cipra Jun 21 '18 at 2:21

$\mathbf{Updated\ 22.06.18}$

Some first values of the sequence $$a_n=\{2^{-1}, 2^{-6^{-1}}, 2^{-6^{-12^{-1}}},\dots 2^{-6^{-12\dots^{{-(n(n+1))^{-1}}}}} \}$$ are $$0.5, 0.890899, 0.550457, 0.867251, 0.56342, 0.860843, 0.566835\dots$$ Easy to see that the even and the odd sequences are different. On the other hand, if the limit $$\lim\limits_{n\to\infty} a_n$$ exists, it must be the limit of the each of the sequences.

Let $$t_n = (n(n+1))^{-((n+1)(n+2))^{-((n+2)(n+3))^{\dots}}},\tag1$$ then $$t_{n} = (n(n+1))^{-t_{n+1}},\tag2$$ $$t_{n+1} = -\dfrac{\log t_{n}}{\log{(n(n+1))}}.\tag3$$

And now let us consider the sequence $T_n,$ such as

$$\lim\limits_{n\to \infty} T_n = \lim\limits_{n\to \infty} T_{n+1},\tag4$$ where $T_n$ is the root of the equation $$T_n = -\dfrac{\log T_n}{\log{(n(n+1))}},\tag5$$ $$T_n = e^{-W(\log(n^2+n))},\tag6$$ where $W(x)$ is the Lambert W-function.

Easy to see that $$2^{-6^{\dots{-((n-1)n)^{-T_n}}}} = 2^{-6^{\dots{-((n-1)n)^{-(n(n+1))^{-T_n}}}}}.\tag7$$ This means that can be defined the sequence $$b_n = 2^{-6^{\dots{-((n-1)n)^{-t_n}}}},\tag8$$ where $$b_1\approx2^{-e^{-W(\log(6))}},$$ $$b_2\approx2^{-6^{-e^{-W(\log(12))}}},$$ $$b_3\approx2^{-6^{-12^{-e^{-W(\log20))}}}}\dots,$$ with more weak difference between the odd/even subsequences.

This approach allows to get more stable estimation of $a$ and supplies the version $a\not=1.$

Numerical calculation for the sequences

Each value of the possible limit $a$ generates a sequence $t_n$ by formulas $(3)$. If the obtained sequence is't monotonic then the value of $a$ is wrong. Consideration of the case $n\to\infty$ allows to get the limits $a_l$ anh $a_h$ for the value of $a.$

For example, the value $a_h=0.719$ generates the sequence $$t_n=\{0.719, 0.475936, 0.414381, 0.354528, 0.311916, 0.311697, 0.289595, 0.289775, 0.275267\},$$ which is not monotonic. Easy to see that sequences with $a>a_h$ are not monotonic too.

This allows to claim that $a<a_h < 0.719.$

Similarly, one can show that $a> a_l > 0.711,$ considering the sequence $$t_n=\{0.711, 0.492079, 0.395766, 0.373025, 0.329171, 0.326702, 0.299306, 0.299673, 0.281777\}$$

Therefore, the possible limit is bounded: $$\boxed{a\in(0.711, 0.719)}.$$

At the same time, numerical calculation for $n=1\dots25$ (step1, step2, step3) shows that the sequence $$t_n \approx \{0.7144, 0.485196, 0.403627, 0.36511, 0.336331, 0.320376, 0.304538, 0.295368, 0.28516, 0.278835, 0.271703, 0.266864, 0.261595, 0.257678, 0.253603, 0.250333, 0.247059, 0.244275, 0.241561, 0.239157, 0.23685, 0.234751, 0.232899, 0.230797, 0.229206\dots\}$$ is monotonic for $n<25.$

On the other hand, if the infinity sequence $t_n,\ n\in 1,2\dots$ for some value $t_1$ is monotonic, then the issue limit exists and $a=t_1.$

Numeric calculation shows that a possible value of the issue limit is $a\approx 0.7144$, if it exists.

  • $\begingroup$ Note that $\frac{0.5677860+0.8588577}{2} \approx 0.71332$. $\endgroup$ – Diger Jun 21 '18 at 2:36
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    $\begingroup$ This argument assumes that the limit exists, i.e., that $t_n$ make sense. However this assumption seems rather unwarranted $\endgroup$ – fedja Jun 21 '18 at 7:27
  • $\begingroup$ @fedja Thanks, fixed. $\endgroup$ – Yuri Negometyanov Jun 21 '18 at 8:33
  • $\begingroup$ @Diger If the limit exists at all. $\endgroup$ – Yuri Negometyanov Jun 21 '18 at 8:34
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    $\begingroup$ Dito. And btw: Stop calling it "consequences" ^^ That awakens completely wrong associations in me.. $\endgroup$ – Diger Jun 22 '18 at 10:47

It is perhaps interesting to note that the behaviour noted by previous posts is true for a wide class of functions defined by towers.

For all positive integers $i$ let $u_i$ be any real numbers such that $1>u_i>0$. Define $$a(n)=u_1^{u_{2}^{...^{u_n}}},b(n)=u_2^{u_{3}^{...^{u_n}}}.$$

Lemma 1 $$a(1)<a(3)<a(5)<a(7)...$$ $$1>a(2)>a(4)>a(6)>a(8)...$$ Proof $$a(N+2)-a(N)=u_1^{b(N+2)}-u_1^{b(N)}.$$ Therefore $a(N+2)-a(N)$ and $b{(N+2)}-b(N)$ have opposite signs.

Now $b(N+2)-b(N)$ is just $a(N+1)-a(N-1)$ for a different sequence and so all the inequalities of the lemma follow inductively from the trivial inequality $1>a(2)$.

Lemma 2

The terms $a(N)$ increase and decrease alternately.

Proof $$a(N+2)-a(N+1)=u_1^{b(N+2)}-u_1^{b(N+1)}.$$ The proof now proceeds similarly to that of Lemma 1.


The $a(2N)$ terms are m.d. to a limit $L$ and the $a(2N+1)$ terms are m.i. to a limit $l$, where $L\ge l$.


This is an immediate consequence of Lemmas 1 and 2 and the fact that the terms $a(N)$ are bounded by $0$ and $1$.


Too long for a comment

The general idea is to interpolate the terms to get a function and then analyze its properties.

Let $\{a_n(x)\}$ be a sequence of once-differentiable functions.

Define the recurrence relation $$A_n(x)=a_n(x)^{A_{n+1}(x)}$$ (where often the '$(x)$' part will be omitted for simplicity.)

Then, we have $$A_n'=A_n\left(A'_{n+1}\ln a_n+A_{n+1}\frac{a_n'}{a_n}\right)$$

Let $$t_n=\frac1n-\frac1{n+1}$$ Let $$H(x)= \begin{cases} 1, &x<0 \\ \frac{\cos(\pi x)+1}2, &0\le x\le1\\ 0, &x>0 \end{cases} $$ Define $$a_n(x)=(t_n)^{H(n-x)}$$

OP’s sequence thus becomes $$\{A_1(1),A_1(2),A_1(3),\cdots\}$$

Then, the limit of the OP's sequence (i.e. $\lim_{n\to\infty}a_{n}$, not to be confused with the $a_n(x)$ in this answer) is $$A_1(\infty)\equiv \lim_{x\to\infty}A_1(x)$$

So our question would become

Does $\lim_{x\to\infty}A_1(x)$ exists?

Let's analyze the derivatives.

Firstly, $$a_n'=-\ln(t_n)H'(n-x)a_n$$ So, $$A_n'=\overbrace{\cdots}^{\text{messy algebra}}=A_nb_n(A_{n+1}H'(n-x)-A'_{n+1}H(n-x))$$ where $b_n=\ln(n(n+1))$.

For $n<\lfloor x\rfloor$, $H'(n-x)=0$. Therefore, we can recursively write out $$A_1'=\left(\prod^{\lfloor x\rfloor}_{k=1}(-A_kb_k)\right) A'_{\lfloor x\rfloor+1}$$

With $$A'_{\lfloor x\rfloor+1}=A_{\lfloor x\rfloor+1}b_{\lfloor x\rfloor+1}(A_{\lfloor x\rfloor+2}H'(\lfloor x\rfloor+1-x)-\underbrace{A'_{\lfloor x\rfloor+2}H(\lfloor x\rfloor+1-x)}_{=0})$$ we can finally write out something neater $$A_1'=-\left(A_{\lfloor x\rfloor+2}\prod^{\lfloor x\rfloor+1}_{k=1}(-A_kb_k)\right)\frac{\sin\pi(x-\lfloor x\rfloor)}2$$

We can easily see the alternation of sign in $A_1’$: whenever $x$ increases one, $A_1’(x)$ changes sign. If the product does not converge to zero, then $A_1’(\infty)\ne0$; and, due to the keep changing of sign, one can expect $A_1(x)$ to keep going up and down as $x$ gets larger and larger. Thus one can argue that the limit $A_1(\infty)$ does not exist.

However, I cannot prove the product does not converge to zero.

  • $\begingroup$ In my opinion to find it's exact limit w'd be an open question $\endgroup$ – zeraoulia rafik Jun 27 '18 at 17:01
  • $\begingroup$ @zeraouliarafik I never attempted to find the exact limit:) $\endgroup$ – Szeto Jun 27 '18 at 22:01
  • $\begingroup$ I shoud upvote to this answer $\endgroup$ – zeraoulia rafik Jun 27 '18 at 22:03
  • $\begingroup$ @zeraouliarafik By the way, what is your view on my answer? $\endgroup$ – Szeto Jun 27 '18 at 22:03

To determine the upper limit (the lower limit can be found similarly)

Let $a_n$ and $u_n$ be as defined in my earlier answer and define $F_n(x)$ to be $${u_{2n-1}^{{u_{2n}}^x}}$$ The following technical result will greatly simplify the working later.

Lemma 3

If $x\ge0.8$ and $u_{2n}\le u_{2n-1}\le0.033$, then $F_n(x)\ge0.8.$


As in Lemma 1, $F_n(x)$ will be minimised (for fixed $u_{2n-1}$) when $x$ is minimised and $u_{2n}$ is maximised.The result therefore follows from the inequality $${t^{t^{0.8}}}\ge 0.8$$ for $t\le 0.033.$

The Inverse function

Since $F_n(x)$ is m.i. on $[0,1]$, it has an inverse. This is given by $$G_n(x)=\frac{\ln(\frac{\ln x}{\ln (u_{2n-1})})}{\ln(u_{2n})}.$$ Then $$a_{2n}=F_1(F_2(...(F_n(1))),1=G_n(...(G_2(G_1(a_{2n})))$$


The limit $L$ for the @zeraoulia rafik sequence satisfies $$0.8588<L<0.8589.$$


Direct calculation of the $a_n$ shows that $L<0.8589.$ Suppose that $L\le 0.8588.$

Another direct calculation shows that $G_7(...(G_2(G_1(0.8588)))<0.8$ and therefore $G_7(...(G_2(G_1(L)))<0.8$.

Lemma 3 applies to the $G_i$ for $i>7$ and so, as $n\to \infty, G_n(...(G_2(G_1(L)))$ does not tend to $1$, a contradiction.

The calculation given in the theorem can, of course, be carried out to whatever degree of precision is required but I have no reason to question the answer provided by @heropup - I am just giving a proof of the previously obtained numerical result with a method which can be used for other tower sequences and can also be adapted to find lower limits.

  • $\begingroup$ Nice answer its seems to build such paper about that problem including you as Author with me , But i will add some analytical results uses dynamical system topic $\endgroup$ – zeraoulia rafik Sep 4 '19 at 14:53
  • $\begingroup$ Great - I look forward to seeing the paper. $\endgroup$ – S. Dolan Sep 5 '19 at 10:53

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