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I have to prove that

$\frac{K_l^2(0)}{K_{l+1}(0) K_{l-1}(0)} = 1 - \frac{1}{l}\quad \text{for} \quad l \ge 2 $

Where $K_l$ is the modified Bessel function of the second kind and $l$ is the order. I'm having trouble figuring out how this gives a non-zero or non-infinite result since $K$ diverges to infinity when $ x \rightarrow 0$. I have looked into expanding it into series but I couldn't end up with anything meaningful, so some help would be appriciated.

sebasket
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1 Answers1

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Use the limiting form from https://dlmf.nist.gov/10.30.E2 $$K_{l}\left(z\right)\sim\tfrac{1}{2}\Gamma\left(l \right)(\tfrac{1}{2}z)^{-l}$$ for $z \to 0$ and get $$\frac{K_l^2(z)}{K_{l+1}(z) K_{l-1}(z)} \sim$$ $$\frac{\Big(\Gamma(l)(\tfrac{1}{2} z)^l\Big)^2} {\Gamma(l-1)(\tfrac{1}{2}z)^{l-1}\cdot\Gamma(l+1)(\tfrac{1}{2}z)^{l+1}}= \frac{\Gamma(l)^2}{\Gamma(l-1)\Gamma(l+1)} = \frac{l-1}{l}=1 -\frac{1}{l}$$

gammatester
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