Convergence problem of $a_n=(1-\frac12)^{(\frac12-\frac13)^{...^{(\frac{1}{n}-\frac{1}{n+1})}}}$
OpenYear of origin: 2001
Posted online: 2021-07-11 01:00:10Z by Rafik Zeraoulia639
Cite as: P-210711.1
Problem's Description
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}}^{......}} $ such that all them value are in $(0,1)$, We want to know if the titled sequence should converge to $1$ ?
Many attempts are given here in SE to show whether the titled limit exist or not ,heropup claimed that 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.In the meanwhile Barry Cipra showed the limit of the titled sequence can't be $ 1$ . The titled sequences assigned the two following Zeraoulia sequencesA328941 and A328942 in OEIS according to the parity of both odd and even iteration. S Dolan stated the following theorem:
Theorem
The limit $L$ for the zeraoulia rafik sequence satisfies $$0.8588< L< 0.8589.$$
One can conclude that the upper and lower bound probably coincide and would be equal yield to proof the existence of limit of the titled sequence. Now according to the given attempts we are ready to reformulate the problem of convergence regarding that sequence as :
Problem: What is the exact limit of $a_n=(1-\frac12)^{(\frac12-\frac13)^{...^{(\frac{1}{n}-\frac{1}{n+1})}}}$, $n\to \infty$ if it exists ?
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Edited: (general update ) at 2021-07-13 17:41:51Z
Created at: 2021-07-11 01:00:10Z View this version
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