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

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.

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.

$\mathbf{Updated\ 22.06.18}$

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.