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Adrien Hardy
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The definition of $I_k$ yields

$$F(\lambda)=e^{-\lambda-1}\sum_{k\geq 1}\sum_{l\geq 0}\frac{\lambda^{k+l}}{l!(k+l)!}=e^{-\lambda-1}\sum_{n\geq 1}a_n\lambda^n$$ where $$a_n=\frac{1}{n!}\sum_{l=0}^{n-1}\frac{1}{l!}.$$ Thus,

$$F'(\lambda)=e^{-\lambda-1}\Big(1+\sum_{n\geq 1 }[(n+1)a_{n+1}-a_n)]\lambda^n\Big)=e^{-\lambda-1}\sum_{n\geq 0}\frac{1}{(n!)^2}\lambda^n$$ and one get $F'(0)$ and $F'(1)$.

NB: I'm pretty late I know, I've just discovered this blog.

Adrien Hardy
  • 2.1k
  • 17
  • 21