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We have a generating function $F_{ij}(s) := \sum_{n=0}^\infty s^n f_{ij}(n)$. Where $f_{ij}(n)$ is a certain probability, so it's in $[0,1]$ for all $n$ (the exact probability that it represents shouldn't be important here).

And we know that $e^{-x} \leq 1-xe^{-x}$ for $x \in [0, \infty)$.

How does this imply that $$ F_{kk}(e^{-x}) \leq 1 - \sum_n nx \cdot f_{kk}(n) \cdot e^{-nx} \quad ? $$

I see that $$ F_{kk}(e^{-x}) \leq F_{kk}(1-xe^{-x}) = \sum_{n=1}^\infty (1-xe^{-x})^n f_{kk}(n), $$ but I don't know how to get the inequality above.

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Start from $$F_{kk}\left(e^{-x}\right) := \sum_{n=0}^\infty \exp\left(-nx\right) f_{kk}(n)\leqslant \sum_{n=0}^\infty\left( 1-nx\exp\left(-nx\right)\right) f_{kk}(n) = \sum_{n=0}^\infty f_{kk}(n)- \sum_{n=0}^\infty nx\exp\left(-nx\right)f_{kk}(n).$$ Then conclude using the fact that $\sum_{n=0}^\infty f_{kk}(n)=1$.

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