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In D. V. Widder, The Laplace Transform, Chapter III, The Moment Problem, given a sequence $(\mu_n)_{n=0}^\infty$, it is defined that $$\lambda_{k,m}:= {k\choose m}(-1)^{k-m}\Delta^{k-m}\mu_m, \, \forall k\ge m,$$ $$\Delta^k\mu_n = \sum_{m=0}^k(-1)^m {k\choose m}\mu_{n+k-m}.$$ Assume $\mu_0>0$, the chapter implies that $$\lambda_{k,m}\ge 0,\,\forall k\ge m\ge0 \Longleftrightarrow \sum_{m=0}^k|\lambda_{k.m}|<L,\, \forall k=0,1,2,\dots, \text{ for some positive }L$$ $(\mu_n)_{n=0}^\infty$ is said to be completely monotonic if it satisfies the condition on the left side above. How do we prove this equivalence, particularly $\Longleftarrow$ algebraically directly without using any integral representation?

It is easily shown that $$\sum_{m=0}^k\lambda_{k.m}=\mu_0,\, \forall k=0,1,2,\dots.$$ So $\implies$ is proved.

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