# Does 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?

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.

• 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 $a_{n}=b_{n-1,n}$. – Sil Jun 12 at 7:26

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.

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.