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Henri Cohen
Henri Cohen
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Thanks to Nemo's suggestion, the answer for $d=0$ is $$\dfrac{a-\delta}{2c}\sum_{n\ge0}\dfrac{n!}{((3+a/\delta)/2)_n}\left(\dfrac{1-a\delta}{2}\right)^n=\dfrac{1}{a-\dfrac{1^2c}{2a-\dfrac{2^2c}{3a-\ddots}}}\;,$$$$\dfrac{a-\delta}{2c}\sum_{n\ge0}\dfrac{n!}{((3+a/\delta)/2)_n}\left(\dfrac{1-a/\delta}{2}\right)^n=\dfrac{1}{a-\dfrac{1^2c}{2a-\dfrac{2^2c}{3a-\ddots}}}\;,$$ with $\delta=\sqrt{a^2-4c}$.

Thanks to Nemo's suggestion, the answer for $d=0$ is $$\dfrac{a-\delta}{2c}\sum_{n\ge0}\dfrac{n!}{((3+a/\delta)/2)_n}\left(\dfrac{1-a\delta}{2}\right)^n=\dfrac{1}{a-\dfrac{1^2c}{2a-\dfrac{2^2c}{3a-\ddots}}}\;,$$ with $\delta=\sqrt{a^2-4c}$.

Thanks to Nemo's suggestion, the answer for $d=0$ is $$\dfrac{a-\delta}{2c}\sum_{n\ge0}\dfrac{n!}{((3+a/\delta)/2)_n}\left(\dfrac{1-a/\delta}{2}\right)^n=\dfrac{1}{a-\dfrac{1^2c}{2a-\dfrac{2^2c}{3a-\ddots}}}\;,$$ with $\delta=\sqrt{a^2-4c}$.

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Henri Cohen
Henri Cohen
  • 13.1k
  • 1
  • 34
  • 62

Thanks to Nemo's suggestion, the answer for $d=0$ is $$\dfrac{a-\delta}{2c}\sum_{n\ge0}\dfrac{n!}{((3+a/\delta)/2)_n}\left(\dfrac{1-a\delta}{2}\right)^n=\dfrac{1}{a-\dfrac{1^2c}{2a-\dfrac{2^2c}{3a-\ddots}}}\;,$$ with $\delta=\sqrt{a^2-4c}$.