Skip to main content
added square root following comment
Source Link
S. Carnahan
  • 45.7k
  • 6
  • 114
  • 220

It looks like you want a formula for the asymptotics of the Pochhammer symbol $(x)_n$ as $n \to \infty$. One such formula is provided about halfway down Wolfram's page:

$$(x)_n \sim \frac{2\pi}{\Gamma(x)} e^{-n}n^{x+n-\frac12}(1+O(\frac1n)) \qquad n \to \infty$$$$(x)_n \sim \frac{\sqrt{2\pi}}{\Gamma(x)} e^{-n}n^{x+n-\frac12}(1+O(\frac1n)) \qquad n \to \infty$$

It looks like you want a formula for the asymptotics of the Pochhammer symbol $(x)_n$ as $n \to \infty$. One such formula is provided about halfway down Wolfram's page:

$$(x)_n \sim \frac{2\pi}{\Gamma(x)} e^{-n}n^{x+n-\frac12}(1+O(\frac1n)) \qquad n \to \infty$$

It looks like you want a formula for the asymptotics of the Pochhammer symbol $(x)_n$ as $n \to \infty$. One such formula is provided about halfway down Wolfram's page:

$$(x)_n \sim \frac{\sqrt{2\pi}}{\Gamma(x)} e^{-n}n^{x+n-\frac12}(1+O(\frac1n)) \qquad n \to \infty$$

Source Link
S. Carnahan
  • 45.7k
  • 6
  • 114
  • 220

It looks like you want a formula for the asymptotics of the Pochhammer symbol $(x)_n$ as $n \to \infty$. One such formula is provided about halfway down Wolfram's page:

$$(x)_n \sim \frac{2\pi}{\Gamma(x)} e^{-n}n^{x+n-\frac12}(1+O(\frac1n)) \qquad n \to \infty$$