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How to prove that $\sum_{k=1}^{\infty}\frac{k^{n+1}}{k!}=eB_{n+1}=1+\cfrac{2^n+\cfrac{3^n+\cfrac{4^n+\cfrac{\vdots}{4}}{3}}{2}}{1}$

score 209 · Answer 1 · 2026-04-17 07:14:43

209

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How to prove that $\sum_{k=1}^{\infty}\frac{k^{n+1}}{k!}=eB_{n+1}=1+\cfrac{2^n+\cfrac{3^n+\cfrac{4^n+\cfrac{\vdots}{4}}{3}}{2}}{1}$

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86

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List Question

How to prove that $\sum_{k=1}^{\infty}\frac{k^{n+1}}{k!}=eB_{n+1}=1+\cfrac{2^n+\cfrac{3^n+\cfrac{4^n+\cfrac{\vdots}{4}}{3}}{2}}{1}$

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