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I try to derive the asymptotic expansion$$a_n=\frac{I(n+1)}{I(n)}=\sqrt{n}+\dfrac{1}{2}-\dfrac{1}{8\sqrt{n}}+o\left(\dfrac{1}{\sqrt{n}}\right)\tag{*}$$from

$$ I(n+1)=\frac{1}{\sqrt{2}} n^{n/2}e^{-n/2+\sqrt{n}-1/4}\left(1+\frac{7}{24\sqrt{n}}+{\cal O}\left(\frac{1}{n}\right)\right). \label{eq1}\tag{1} $$

It is given in the end of this answer

$$ I(n+1)=\frac{1}{\sqrt{2}} n^{n/2}e^{-n/2+\sqrt{n}-1/4}\left(1+\frac{7}{24\sqrt{n}}+{\cal O}\left(\frac{1}{n^{3/4}}\right)\right). $$ Now, it is a "simple" matter to conclude from this that $a_n$ has the desired asymptotic expansion.
Edit: In fact the term $\mathcal{O}(n^{-3/4})$ effectively destroys the asymptotic expansion as mercio noted, But in fact we have $$\tag1 I(n+1)=\frac{1}{\sqrt{2}} n^{n/2}e^{-n/2+\sqrt{n}-1/4}\left(1+\frac{7}{24\sqrt{n}}+{\cal O}\left(\frac{1}{n}\right)\right). $$

My working:

$$a_n=\frac{I(n+1)}{I(n)}=\frac{n^{n/2}e^{-n/2+\sqrt{n}}}{(n-1)^{(n-1)/2}e^{-(n-1)/2+\sqrt{n-1}}}\frac{\left(1+\frac{7}{24\sqrt{n}}+{\cal O}\left(\frac{1}{n}\right)\right)}{\left(1+\frac{7}{24\sqrt{n-1}}+{\cal O}\left(\frac{1}{n}\right)\right)}$$ Since $\frac1{\sqrt{n-1}}=\frac1{\sqrt{n}}+{\cal O}\left(\frac{1}{n\sqrt{n}}\right)$, I get $\left(1+\frac{7}{24\sqrt{n-1}}+{\cal O}\left(\frac{1}{n}\right)\right)=\left(1+\frac{7}{24\sqrt{n}}+{\cal O}\left(\frac{1}{n}\right)\right)$, so$$\frac{\left(1+\frac{7}{24\sqrt{n}}+{\cal O}\left(\frac{1}{n}\right)\right)}{\left(1+\frac{7}{24\sqrt{n-1}}+{\cal O}\left(\frac{1}{n}\right)\right)}=\frac{\left(1+\frac{7}{24\sqrt{n}}+{\cal O}\left(\frac{1}{n}\right)\right)}{\left(1+\frac{7}{24\sqrt{n}}+{\cal O}\left(\frac{1}{n}\right)\right)}=1+{\cal O}\left(\frac{1}{n}\right)$$ so $$a_n=\frac{n^{n/2}e^{-n/2+\sqrt{n}}}{(n-1)^{(n-1)/2}e^{-(n-1)/2+\sqrt{n-1}}}\left(1+{\cal O}\left(\frac{1}{n}\right)\right)$$ Since $\sqrt{n}-\sqrt{n-1}=\frac12n^{-1/2}+{\cal O}(n^{-3/2})$, I get $\frac{e^{-n/2+\sqrt{n}}}{e^{-(n-1)/2+\sqrt{n-1}}}=e^{-1/2+\sqrt{n}-\sqrt{n-1}}=e^{-1/2}(1+\frac12n^{-1/2}+{\cal O}(n^{-1}))$
so $$a_n=\frac{n^{n/2}}{(n-1)^{(n-1)/2}}e^{-1/2}\left(1+\frac12n^{-1/2}+{\cal O}(n^{-1})\right)\\=\frac{n^{n/2}}{(n-1)^{n/2}}(n-1)^{1/2}e^{-1/2}\left(1+\frac12n^{-1/2}+{\cal O}(n^{-1})\right)$$ Since $(n-1)^{1/2}=\sqrt{n}(1+{\cal O}(n^{-1}))$, I get $$a_n=\frac{n^{n/2}}{(n-1)^{n/2}}\sqrt{n}e^{-1/2}\left(1+\frac12n^{-1/2}+{\cal O}(n^{-1})\right)\\ =(1-\frac1n)^{-n/2}\sqrt{n}e^{-1/2}\left(1+\frac12n^{-1/2}+{\cal O}(n^{-1})\right)$$ Since $(1-\frac1n)^{-n/2}=e^{-\frac n2\log(1-\frac1n)}=e^{-\frac n2(-n^{-1}+{\cal O}(n^{-2}))}=e^{\frac12+{\cal O}(n^{-1})}=e^{\frac12}+{\cal O}(n^{-1})$, then I get $$a_n=\sqrt{n}\left(1+\frac12n^{-1/2}+{\cal O}(n^{-1})\right)=\sqrt{n}+\frac12+{\cal O}(n^{-1/2})$$

but this is less precise than $(\text*)$.

I doubt that I cannot get the $-\frac{1}{8\sqrt{n}}$ term from \eqref{eq1} because ${\cal O}(\frac1n)$ appear in \eqref{eq1}.

The $\sqrt{n}$ term in $(\text*)$, when multiplied by ${\cal O}(\frac1n)$, becomes ${\cal O}(\frac1{\sqrt{n}})$ and absorbs the term $-\frac1{8\sqrt{n}}$.

so I need to find one more term in \eqref{eq1}, change ${\cal O}(\frac1{n})$ to $o(\frac1{n})$ in order to get the $-\frac{1}{8\sqrt{n}}$ term, is this correct?

hbghlyj
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  • I think you are right. From the paper linked in Omran Kouba, I belive it follows that $I(n+1)$ has an asymptotic expansion in negative powers of $\sqrt{n}$. From that it follows that $a_n$ itself has such an asymptotic expansion. The coefficients may be calculated by substituting the expansion into the recurrence for $a_n$, expanding everything in $\sqrt{n}$, and comparing the coefficients of like-powers of $\sqrt{n}$. – Gary Apr 09 '24 at 23:39
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    See also https://math.stackexchange.com/q/3943534 – Gary Apr 10 '24 at 03:46
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    The error term is correct. Moser and Wyman (eq. 3.40) showed that $$ I(n + 1) \sim \frac{1}{{\sqrt 2 }}n^{n/2} {\rm e}^{ - n/2 + \sqrt n - 1/4} \left( {1 + \frac{7}{{24\sqrt n }} - \frac{{119}}{{1152n}} + \ldots } \right) $$ as $n\to+\infty$. – Gary Apr 10 '24 at 03:52
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    May be, my answer at https://math.stackexchange.com/questions/3942653/given-the-sequence-a-1-1-a-n1-1-fracna-n-does-the-sequence-a-nn-a/4899469#4899469 could be of interest to you. – Claude Leibovici Apr 15 '24 at 10:56

1 Answers1

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Using $I(n+1)=\frac{1}{\sqrt{2}} n^{n/2}e^{-n/2+\sqrt{n}-1/4}\left(1+\frac{7}{24\sqrt{n}}+{\cal O}\left(\frac{1}{n}\right)\right)$, we cannot get the $-\frac1{8\sqrt{n}}$ term

SageMath code:

A.<n> = AsymptoticRing(growth_group='n^QQ * log(n)^ZZ', coefficient_ring=QQ)
f = log(n)*(n/2) + (-n/2 + sqrt(n)) + log(1 + 7/(24*sqrt(n)) + O(1/n))
e^(f - f.subs(n=n-1))

Output: n^(1/2) + 1/2 + O(n^(-1/2))

To get the $-\frac1{8\sqrt{n}}$ term, we need include more term in $I(n+1)$, increase the order ${\cal O}\left(\frac{1}{n}\right)$ to ${\cal O}\left(\frac1{n^{3/2}}\right)$

Using $I(n + 1) = \frac{1}{{\sqrt 2 }}n^{n/2} {\rm e}^{ - n/2 + \sqrt n - 1/4} \left( 1 + \frac{7}{{24\sqrt n }} - \frac{{119}}{{1152n}} + {\cal O}\left(\frac1{n^{3/2}}\right)\right)$ from @Gary's comment

A.<n> = AsymptoticRing(growth_group='n^QQ * log(n)^ZZ', coefficient_ring=QQ)
f = log(n)*(n/2) + (-n/2 + sqrt(n)) + log(1 + 7/(24*sqrt(n)) - 119/(1152*n) + O(1/n^(3/2)))
e^(f - f.subs(n=n-1))

Output: n^(1/2) + 1/2 - 1/8*n^(-1/2) + O(n^(-1))

The effect of including the term $- \frac{{119}}{{1152n}}$ in the computation is changing $$\frac{1+\frac{7}{24\sqrt{n}}+{\cal O}\left(\frac{1}{n}\right)}{1+\frac{7}{24\sqrt{n-1}}+{\cal O}\left(\frac{1}{n}\right)}=\frac{1+\frac{7}{24\sqrt{n}}+{\cal O}\left(\frac{1}{n}\right)}{1+\frac{7}{24\sqrt{n}}+{\cal O}\left(\frac{1}{n}\right)}=1+{\cal O}\left(\frac{1}{n}\right)$$

to $$\frac{1+\frac{7}{24\sqrt{n}}- \frac{{119}}{{1152n}}+{\cal O}\left(\frac{1}{n^{3/2}}\right)}{1+\frac{7}{24\sqrt{n-1}}- \frac{{119}}{{1152(n-1)}}+{\cal O}\left(\frac{1}{n^{3/2}}\right)}=\frac{1+\frac{7}{24\sqrt{n}}- \frac{{119}}{{1152n}}+{\cal O}\left(\frac{1}{n^{3/2}}\right)}{1+\frac{7}{24\sqrt{n}}- \frac{{119}}{{1152n}}+{\cal O}\left(\frac{1}{n^{3/2}}\right)}=1+{\cal O}\left(\frac{1}{n^{3/2}}\right)$$

so the coefficients $\frac7{24}$ and $-\frac{119}{1152}$ don't matter in this calculation to get $1+{\cal O}\left(\frac{1}{n^{3/2}}\right)$.

Gary
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hbghlyj
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    Indeed, $$\frac{{1 + \frac{a}{{\sqrt n }} + \frac{b}{n} + \mathcal{O}!\left( {\frac{1}{{n^{3/2} }}} \right)}}{{1 + \frac{a}{{\sqrt {n - 1} }} + \frac{b}{{n - 1}} + \mathcal{O}!\left( {\frac{1}{{(n - 1)^{3/2} }}} \right)}} = 1 + \mathcal{O}!\left( {\frac{1}{{n^{3/2} }}} \right) $$ for any complex $a$ and $b$. – Gary Apr 11 '24 at 00:37