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

Solution 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.

Solution 2:

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.

Solution 3:

$\mathbf{Updated\ 22.06.18}$

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.

Solution 4:

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.

Solution 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 $1$.