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, 2018 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? $\endgroup$ Jun 16, 2018 at 23:13
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    $\begingroup$ No, it is my own problem , The motivation is to look what about telescopic using power $\endgroup$ Jun 16, 2018 at 23:14
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    $\begingroup$ @user625 Nope, your recursion is wrong. Your recursion would give the sequence $$a_n=(1/2)^{(1/6)(1/12)(1/20)...(1/n(n+1))}$$ $\endgroup$ Jun 16, 2018 at 23:24
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    $\begingroup$ @Frpzzd you are right. $\endgroup$
    – user625
    Jun 16, 2018 at 23:41

7 Answers 7


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, 2018 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, 2018 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, 2018 at 13:05
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    $\begingroup$ That sequence is already assigned A328941 and A328942 $\endgroup$ Nov 1, 2019 at 16:03
  • $\begingroup$ @Szeto: not a plot but a list of convergence up to $n$ and $n+1$ steps $$\small \begin{bmatrix} 1 & 0.50000000 & 0.89089872 \\ 3 & 0.55045661 & 0.86725125 \\ 5 & 0.56342048 & 0.86084299 \\ 7 & 0.56683473 & 0.85925354 \\ 9 & 0.56761318 & 0.85892332 \\ 11 & 0.56775989 & 0.85886682 \\ 13 & 0.56778272 & 0.85885879 \\ 15 & 0.56778570 & 0.85885783 \\ 17 & 0.56778603 & 0.85885773 \\ 19 & 0.56778606 & 0.85885772 \\ 21 & 0.56778607 & 0.85885772 \\ 23 & 0.56778607 & 0.85885772 \end{bmatrix}$$ $\endgroup$ Jan 6 at 22:06

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, 2018 at 1:38
  • $\begingroup$ @Diger, yes, you're right. I'll fix it. Thank you! $\endgroup$ Jun 21, 2018 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, 2018 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, 2018 at 7:27
  • $\begingroup$ @fedja Thanks, fixed. $\endgroup$ Jun 21, 2018 at 8:33
  • $\begingroup$ @Diger If the limit exists at all. $\endgroup$ Jun 21, 2018 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, 2018 at 10:47

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 $\endgroup$ Jun 27, 2018 at 17:01
  • $\begingroup$ @zeraouliarafik I never attempted to find the exact limit:) $\endgroup$
    – Szeto
    Jun 27, 2018 at 22:01
  • $\begingroup$ I shoud upvote to this answer $\endgroup$ Jun 27, 2018 at 22:03
  • $\begingroup$ @zeraouliarafik By the way, what is your view on my answer? $\endgroup$
    – Szeto
    Jun 27, 2018 at 22:03

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 $1$.


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 $\endgroup$ Sep 4, 2019 at 14:53
  • $\begingroup$ Great - I look forward to seeing the paper. $\endgroup$
    – user502266
    Sep 5, 2019 at 10:53

Note: every number $M$ has a number between $0$ and $1$ and we have $a \leq b\leq c$ then $M^{c}\leq M^{b}\leq M^{a}$\

\begin{equation*} (\dfrac{1}{2})^{(\dfrac{1}{2(1)})^{(\dfrac{1}{2(1)})^{...^{(\dfrac{1}{2(1)})}}}} \leq (\dfrac{1}{2})^{(\dfrac{1}{2(1+2)})^{(\dfrac{1}{2(1+2+3)})^{...^{(\dfrac{1}{2(1+2+3+4+...+n)})}}}} \end{equation*} \begin{equation} \label{1} (\dfrac{1}{2})^{(\dfrac{1}{2})^{(\dfrac{1}{2})^{...^{(\dfrac{1}{2})}}}} \leq (\dfrac{1}{2})^{(\dfrac{1}{2(1+2)})^{(\dfrac{1}{2(1+2+3)})^{...^{(\dfrac{1}{2(1+2+3+4+...+n)})}}}} \end{equation} we've got
$ (1+2) < (1\times2\times3) = 3! < n! $
$ (1+2+3) < (1\times2\times3)= 3! < n! $
$ (1+2+3+4)< (1\times2\times3\times 4) =4! < n! $
$\qquad\qquad\qquad\vdots \qquad\qquad\qquad\qquad\qquad\qquad \vdots\qquad\vdots$
$ (1+2+3+4+\cdots+n)< (1\times2\times3\times\cdots n) =n! = n! $
so \begin{equation*} (\dfrac{1}{2})^{(\dfrac{1}{2(1+2)})^{(\dfrac{1}{2(1+2+3)})^{...^{(\dfrac{1}{2(1+2+3+4+...+n)})}}}}\leq (\dfrac{1}{2})^{(\dfrac{1}{2\times n!})^{(\dfrac{1}{2\times n!})^{...^{(\dfrac{1}{2\times n!})}}}} \end{equation*} The larger the denominator, the smaller the number
we have $ n!\leq n!\times n! $ \begin{equation} \label{2} (\dfrac{1}{2})^{(\dfrac{1}{2(1+2)})^{(\dfrac{1}{2(1+2+3)})^{...^{(\dfrac{1}{2(1+2+3+4+...+n)})}}}}\leq (\dfrac{1}{2})^{(\dfrac{1}{2\times n!\times n!})^{(\dfrac{1}{2\times n!\times n!})^{...^{(\dfrac{1}{2\times n!\times n!})}}}} \end{equation} So from inequality 1 and inequality 2 we find \begin{equation*} \label{3}(\dfrac{1}{2})^{(\dfrac{1}{2})^{(\dfrac{1}{2})^{...^{(\dfrac{1}{2})}}}} \leq (\dfrac{1}{2})^{(\dfrac{1}{2(1+2)})^{(\dfrac{1}{2(1+2+3)})^{...^{(\dfrac{1}{2(1+2+3+4+...+n)})}}}}\leq \end{equation*} \begin{equation} \leq (\dfrac{1}{2})^{(\dfrac{1}{2\times n!\times n!})^{(\dfrac{1}{2\times n!\times n!})^{...^{(\dfrac{1}{2\times n!\times n!})}}}} \end{equation} We use the Lim ,we have
\begin{equation*} \label{4}\displaystyle\lim_{n\rightarrow +\infty}(\dfrac{1}{2})^{(\dfrac{1}{2})^{(\dfrac{1}{2})^{...^{(\dfrac{1}{2})}}}} \leq \lim_{n\rightarrow +\infty}(\dfrac{1}{2})^{(\dfrac{1}{2(1+2)})^{(\dfrac{1}{2(1+2+3)})^{...^{(\dfrac{1}{2(1+2+3+4+...+n)})}}}} \end{equation*} \begin{equation}\leq \lim_{n\rightarrow +\infty}(\dfrac{1}{2})^{(\dfrac{1}{2\times n!\times n!})^{(\dfrac{1}{2\times n!\times n!})^{...^{(\dfrac{1}{2\times n!\times n!})}}}} \end{equation} remarque
$ \lim_{n\rightarrow+\infty}\dfrac{1}{2\times n!\times n! }=\lim_{n\rightarrow+\infty}\dfrac{1}{2}\times \dfrac{1}{ n!}\times\dfrac{1}{n!} =\lim_{n\rightarrow+\infty}\dfrac{1}{2}\times\dfrac{1}{\dfrac{ n!}{n!} }=\dfrac{1}{2} $
NOte $ \lim_{n\rightarrow+\infty}\dfrac{ n!}{n!}=1 $ \begin{equation*} \label{5}\displaystyle\lim_{n\rightarrow +\infty}(\dfrac{1}{2})^{(\dfrac{1}{2})^{(\dfrac{1}{2})^{...^{(\dfrac{1}{2})}}}} \leq \lim_{n\rightarrow +\infty}(\dfrac{1}{2})^{(\dfrac{1}{2(1+2)})^{(\dfrac{1}{2(1+2+3)})^{...^{(\dfrac{1}{2(1+2+3+4+...+n)})}}}}\leq \end{equation*} \begin{equation} \leq\lim_{n\rightarrow +\infty} (\dfrac{1}{2})^{(\dfrac{1}{2\times\dfrac{ n!}{ n!} })^{(\dfrac{1}{2\times\dfrac{ n!}{ n!}})^{...^{(\dfrac{1}{2\times\dfrac{ n!}{ n!}})}}}} \end{equation} \begin{equation*} \label{6}(\dfrac{1}{2})^{(\dfrac{1}{2})^{(\dfrac{1}{2})^{...^{(\dfrac{1}{2})}}}} \leq\lim_{n\rightarrow +\infty}(\dfrac{1}{2})^{(\dfrac{1}{2(1+2)})^{(\dfrac{1}{2(1+2+3)})^{...^{(\dfrac{1}{2(1+2+3+4+...+n)})}}}}\leq (\dfrac{1}{2})^{(\dfrac{1}{2})^{(\dfrac{1}{2})^{...^{(\dfrac{1}{2})}}}} \end{equation*} so
$\lim_{n\rightarrow +\infty}a_{n}=(\dfrac{1}{2})^{(\dfrac{1}{2})^{(\frac{1}{2})^{...^{(\dfrac{1}{2})}}}}\simeq 0.6411857445$


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