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I have found this limit in https://oeis.org/A019609 and I was wondering how to prove it (if it is actually correct): $$\lim_{n\to\infty} \frac{4n}{a^2_n}=\pi e$$ where $$a_1=0,a_2=1, a_{n+2}=a_{n+1}+\frac{a_n}{2n}.$$

By computer evaluation, it is correct for $2$ digits after decimal point at about $n\approx 24100$, so if it is correct, it converges really slow.

I've attempted to prove this by first considering generating function $f(x)=\sum_{n \geq 1}a_nx^n$ and then trying to get asymptotics of its coefficients. By using recurrence, we get $f(x)/x^2-1=f(x)/x+\sum \frac{a_n}{2n}x^n$, and after differentiation we get differential equation which solves to $$f(x)=\frac{e^{-x/2}x^2}{(1-x)^{3/2}}.$$ Now I think this is a step away from getting asymptotics of $a_n$, but I don't know how. Can anybody show how to finish this? Or maybe there is another way?

Also, I don't think it is useful, but here is at least closed form obtained from the $f(x)$ using binomial series and exponential function series: $$ a_n=\sum_{i=0}^{n-2}\frac{(-1)^n}{2^i i!}\binom{-3/2}{n-i-2}. $$

Closest to this question seems to be Mirror algorithm for computing $\pi$ and $e$ - does it hint on some connection between them?, where there are two sequences approaching $\pi$ and $e$ and solutions seem to use same approach using generating functions, so this seems to be on the right track.

Sil
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  • I've just realized that one of the answers in referenced question gives asymptotic for generic case $u_{n+2} = a u_{n+1} + b\frac{u_n}n$, I guess then it becomes easy... – Sil Apr 21 '20 at 19:41

2 Answers2

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$$a_n=\sum_{i=0}^{n-2}\frac{(-1)^n}{2^i i!}\binom{-3/2}{n-i-2}=(-1)^n \binom{-\frac{3}{2}}{n-2} \, _1F_1\left(2-n;\frac{3}{2}-n;-\frac{1}{2}\right)$$ $$\frac {4n}{a_n^2}=\frac{4 n}{\binom{-\frac{3}{2}}{n-2}^2 \,\,\Big[\, _1F_1\left(2-n;\frac{3}{2}-n;-\frac{1}{2}\right){}\Big]^2}$$

For large values of $n$ $$\frac{4 n}{\binom{-\frac{3}{2}}{n-2}^2}=\pi +\frac{5 \pi }{4 n}+\frac{49 \pi }{32 n^2}+O\left(\frac{1}{n^3}\right)$$

Now, computing the values of the Kummer's confluent hypergeometric function, the numerators are $$\{0,1,2,10,68,604,6584,85048,1269680,21505552\}$$ and they correspond to sequence $A165968$ in $OEIS$. For this, Vaclav Kotesovec proposed as an approximation $$2^{(n-\frac{1}{2})} e^{-(n+\frac{1}{2})} n^{(n-1)}$$

The denominators are the double factorials of odd numbers.

Using Vaclav Kotesovec' approximation for the numerator we end with $$\, _1F_1\left(2-n;\frac{3}{2}-n;-\frac{1}{2}\right)\sim \frac{2^{(n-\frac{1}{2})} e^{-(n+\frac{1}{2})} n^{(n-1)}}{(2n-3)!!}$$

Using Stirling like approximation for the denominator, $$\, _1F_1\left(2-n;\frac{3}{2}-n;-\frac{1}{2}\right)\sim \frac{1}{\sqrt{e}}-\frac{11}{24 \sqrt{e} n}-\frac{23}{1152 \sqrt{e} n^2}+O\left(\frac{1}{n^3}\right)$$ which finally makes $$\frac {4n}{a_n^2}=e\pi \left(1+\frac{13}{6 n}+\frac{241}{72 n^2} \right)+O\left(\frac{1}{n^3}\right)$$

Edit

I made a mistake trying to get the asymptotics of Kummer's function from its pieces.

Using what @Gary gave in comments, we end with $$\frac {4n}{a_n^2}=e\pi \left(1+\frac{3}{4 n}+\frac{13}{32 n^2}+\frac{1}{128 n^3}-\frac{669}{2048 n^4} \right)+O\left(\frac{1}{n^5}\right)$$ which is very different even for the expansion up to $O\left(\frac{1}{n^2}\right)$.

Claude Leibovici
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    To be sure about this result, you need to know that the difference between ${}_1 F_1$ and Vaclav Kotesovec's approximation is at most $\mathcal{O}(n^{-3})$. – Gary Apr 22 '20 at 06:25
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    The correct expansion is as follows: \begin{align} & {}1F_1 \big( {2 - n;\tfrac{3}{2} - n; - \tfrac{1}{2}} \big) = \frac{1}{{\sqrt e }}{}_1F_1 \big( { - \tfrac{1}{2};\tfrac{3}{2} - n;\tfrac{1}{2}} \big) \ & \sim \frac{1}{{\sqrt e }}\sum\limits{k = 0}^\infty {\binom{2k}{k}\frac{{( - 1)^k }}{{4^k (1 - 2k)}}\frac{1}{{(2n - 3)(2n - 5) \cdots (2n - (2k + 1))}}} \ & \sim \frac{1}{{\sqrt e }}\left( {1 + \frac{1}{{4n}} + \frac{{11}}{{32n^2 }} + \frac{{57}}{{128n^3 }} + \frac{{1059}}{{2048n^4 }} + \cdots } \right). \end{align} (cf. http://dlmf.nist.gov/13.8.E1) – Gary Apr 22 '20 at 07:01
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You can get the asymptotics of the coefficients of the generating function:

$$f(z)=\frac{e^{-z/2} z^2}{(1-z)^{3/2}}$$

using standard tools of singularity analysis from analytic combinatorics (see e.g. section B.VI of Flagolet and Sedgewick's book). What you need is:

$$[z^n](1-z)^{-\alpha}\underset{n\to\infty}{\sim} \frac{n^{\alpha-1}}{\Gamma(\alpha)}$$

And some transfer theorem, namely that under mild conditions on the regularity of $f$ on the unit disk (satisfied here), $f(z)\underset{z\to 1}{\sim} C(1-z)^{-\alpha}$ implies that $[z^n] f(z)\underset{n\to\infty}{\sim} Cn^{\alpha-1}/\Gamma(\alpha)$. Basically it allows you to say directly:

$$[z^n] f(z)\underset{n\to\infty}{\sim} e^{-1/2}\frac{\sqrt{n}}{\Gamma(3/2)}=2\sqrt{\frac{n}{e\pi}}$$

which gives the intended asymptotics.

md5
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  • Nice! I guess the $C=e^{-1/2}$ comes from unit disc radius $x=1$ in that $e^{-x/2}$ expression? Will have to study the reference. – Sil Apr 21 '20 at 20:08
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    @Sil: Yes, that's it. Basically the idea is that an equivalent of $f$ around its first singularity will directly give you an equivalent of $[z^n] f(z)$ as $n\to \infty$. So anytime you need to do such a computation, you only have to check in a "dictionary" how do functions from your asymptotic scale in the complex plane convert in the "coefficients world" – md5 Apr 21 '20 at 20:22