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Does the following limit exists?

$$\lim_{n \rightarrow \infty} \frac{p(n)}{p(n-5)}$$

where $p(n)$ denote the partition function.

If this limit exists, is it equal to 1?

Kindly share your thoughts.

Thank you.

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Using $$p(n) \sim \frac{e^{\pi\sqrt{\frac{2n}3}}}{4\sqrt3\,n} $$ compute $\frac{p(n)}{p(n-a)}$, take logarithms and expand as series for large values og $n$. This would give $$\log \left(\frac{p(n)}{p(n-a)}\right)=\frac{\pi a}{\sqrt{6n}}-\frac{a}{2 n}+O\left(\frac{1}{n^{3/2}}\right)$$ and continuing with Taylor series using $x=e^{\log(x)}$, then $$\frac{p(n)}{p(n-a)}=1+\frac{\pi a }{\sqrt{6n}}+\frac{a \left(\pi ^2 a-6\right)}{12 n}+O\left(\frac{1}{n^{3/2}}\right)$$ which shows the limit (already given in answers) but also a quite good approximation of the ratio.

Using $n=50$ and $a=5$, the exact value is $\frac{102113}{44567}\approx 2.291$ while the above expansion would give $\frac{114+20 \sqrt{3} \pi +5 \pi ^2}{120} \approx 2.268$

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  • $\begingroup$ Thank you. Very insightful. $\endgroup$ – GA316 Jan 3 at 8:30
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$p(n) \sim \frac1{4\sqrt3n} e^{\pi\sqrt{2n/3}}$ according to OEIS and GTM177 Chapter II.
Hence, $$\lim_{n \rightarrow \infty} \frac{p(n)}{p(n-5)}=\lim_{n \rightarrow \infty} \frac{\frac1{4\sqrt3n} e^{\pi\sqrt{2n/3}}}{\frac1{4\sqrt3(n-5)} e^{\pi\sqrt{2(n-5)/3}}}=1$$

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