Convergence of $a_n=(1-\frac12)^{(\frac12-\frac13)^{...^{(\frac{1}{n}-\frac{1}{n+1})}}}$

Question

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 $\infty$ ?

Answer 1

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.

Answer 2

$\mathbf{Updated\ 22.06.18}$

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 $a\not=1.$

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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.

Answer 3

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

$$0\lt(1/12)\lt(1/12)^{(1/20)^\cdots}\lt1$$

and thus also that

$$0\lt(1/6)\lt(1/6)^{(1/12)^{(1/20)^\cdots}}\lt(1/6)^{(1/12)}\lt1$$

so that, finally,

$$0.5504566141\approx(1/2)^{(1/6)^{(1/12)}}\lt(1/2)^{(1/6)^{(1/12)^\cdots}}\lt(1/2)^{(1/6)}\approx0.89089871814$$

These bounds accord with what heropup found.

Answer 4

Too long for a comment

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

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

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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.

Answer 5

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.

Theorem

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 $L\ge l$.

Proof

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

Answer 6

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 $F_n(x)\ge0.8.$

Proof

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 $$a_{2n}=F_1(F_2(...(F_n(1))),1=G_n(...(G_2(G_1(a_{2n})))$$

Theorem

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

Proof

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.

Answer 7

Blockquote
$a_{n}=(1-\dfrac{1}{2})^{(\dfrac{1}{2}-\dfrac{1}{3})^{(\dfrac{1}{3}-\dfrac{1}{4})^{...^{(\dfrac{1}{n}-\dfrac{1}{n+1})}}}}$
$a_{n}=(1-\dfrac{1}{2})^{(\dfrac{1}{6})^{(\dfrac{1}{12})^{...^{(\dfrac{1}{n(n+1)})}}}}$
$a_{n}=(\dfrac{1}{2})^{(\dfrac{1}{2(1+2)})^{(\dfrac{1}{2(1+2+3)})^{...^{(\dfrac{1}{2(1+2+3+4+...+n)})}}}}$
Note: every number $M$ has a number between $0$ and $1$ and we have $a \leq b\leq c$ then $M^{c}\leq M^{b}\leq M^{a}$\

Which
\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
$\lim_{n\rightarrow +\infty}a_{n}=(\dfrac{1}{2})^{(\dfrac{1}{2})^{(\frac{1}{2})^{...^{(\dfrac{1}{2})}}}}\simeq 0.6411857445$

Answer 8

$\forall n\in N^{*}:u_{n}=(1-\frac{1}{2})^{(\frac{1}{2}-\frac{1}{3})^{(\frac{1}{3}-\frac{1}{4})^{...(\frac{1}{n}-\frac{1}{n+1})}}}\gt 0$

$u_{1}=v_{1}=1-\frac{1}{2}=\frac{1}{2}$

$u_{2}=v_{1}^{v_{2}},u_{3}=v_{1}^{v_{2}^{v_{3}}},...,u_{n}=v_{1}^{v_{2}^{v_{3}^{....v_{n}}}},v_{n}=\frac{1}{n}-\frac{1}{n+1}=\frac{1}{n(n+1)}$

$\ln u_{n}=v_{2}^{v_{3}^{^{....v_{n}}}}\ln v_{1}=-v_{2}^{v_{3}^{^{....v_{n}}}}\ln 2\lt 0$

$0\lt u_{n}\lt 1$

$\frac{1}{9900}^{\frac{1}{10100}}\approx 0.9991$

$\lim_{n \to \infty }(\frac{1}{(n-1)n})^{\frac{1}{n(n+1)}}=1 $

$n\ge 99\Rightarrow v_{n}^{v_{n+1}}\gt 0.999$

$ A_{n}=v_{99}^{v100^{v_{101}^{...v_{n}}}}\longrightarrow \lim_{n \to \infty }A_{n}\approx 1$

$\lim_{n \to \infty }u_{n}=v_{1}^{v_{2}^{^{...v_{99}}}} $