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My question here is related to telescopic sum using factorial and it is related to my question here, I have computed some values of $a_n=(1-\frac{1}{2!})^{(\frac{1}{2!}-\frac{1}{3!})^{\ldots^{(\frac{1}{n!}-\frac{1}{(n+1)!})}}}$ for odd parity and even parity but it is not fixed for example for $n=2$ we have $0.793700$ and it decreases for $n=4$ to $0.77982$, now for $n=3$ we have $0.5465$ and it increases for $n=5$ to $0.54876$ , it seems increasing for odd parity and decreasing for even parity iteration. Now I have looked to all given answers here but I can't juge whether that sequence converges or not by means it has a limit or not?

My question here is: Is this $a_n=(1-\frac{1}{2!})^{(\frac{1}{2!}-\frac{1}{3!})^{\ldots^{(\frac{1}{n!}-\frac{1}{(n+1)!})}}}$ have a finit limit ?

Note The motivation of this question is looking to the behavior of the Gamma function in the power telescoping sum.

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  • $\begingroup$ It might be (or might not be at all) useful to state it recursively by $b_{0,m}=\frac{1}{m!}-\frac{1}{(m+1)!}$, $b_{n,m}=(\frac{1}{(m-n)!}-\frac{1}{(m+1-n)!})^{b_{n-1,m}}$, then $a_{n}=b_{n-1,n}$. $\endgroup$ – Sil Jun 12 at 7:26
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We have similar behavior to the sequence that you linked to, except the limiting values are different: For even $n$, it is $$a_n \to 0.77954333600168773503298455024204190801488463615921\ldots,$$ and for odd $n$, it is $$a_n \to 0.54877354704085687513069922740691455562600046738030\ldots.$$ The number of correct decimal places increases slightly faster than quadratically in $n$; i.e., if $\epsilon(n)$ is the absolute error as a function of $n$, then $$-\log \epsilon(n) \sim O(n^{k}),$$ where I estimate $k$ to be approximately $2.3$, certainly greater than $2$ but less than $2.4$. I do not recognize these constants as having a closed form.

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I don't think this really counts as a proof, but I made some code in Matlab code to investigate this.

n=input('please input a value for n ');

for m=1:n;

for I=1:m;

    b(m+1)=1;

    b(m-I+1)=(1/factorial(m-I+1)-1/factorial(m-I+2))^b(m-I+2);

    a(m)=b(1);

end

end

for I=1:n;

fprintf('a(%.f)=%.8f \n',I,a(I));

end

Inputting n=12 outputs a(1)=0.50000000 a(2)=0.79370053 a(3)=0.54650798 a(4)=0.77982290 a(5)=0.54876028 a(6)=0.77954370 a(7)=0.54877354 a(8)=0.77954334 a(9)=0.54877355 a(10)=0.77954334 a(11)=0.54877355 a(12)=0.77954334 So while it seems that $a_{2n}$ and $a_{2n-1}$ each converge, they don't converge to the sam value.

The thing is that as n tends to infinity $(\frac{1}{(n)!}-\frac{1}{(n+1)!})$ rapidly tends to 0, so $(\frac{1}{(m-n)!}-\frac{1}{(m+1-n)!})^{(\frac{1}{(n+1)!}-\frac{1}{(n+2)!})}$ quickly tends to 1.

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