“Chaplin or Leibniz ?” Revisit

Once the closed form of power summation is derived (see “Little Bird and a Recursive Generator“) namely

 s_{p} = 1^p+2^p+3^p+\dots+n^p=\frac{n(n+1)^p - \sum\limits_{j=2}^{p}\binom{p}{j} s_{p-j+1}}{p+1}, \quad\quad p, n \in N^{+},

the validity of the inequality in “Chaplin or Leibniz ?” is readily shown:

 1^p+2^p+3^p+\dots+n^p = \frac{n(n+1)^p - \sum\limits_{j=2}^{p}\binom{p}{j} s_{p-j+1}}{p+1}

< \frac{(n+1)(n+1)^p - \sum\limits_{j=2}^{p}\binom{p}{j} s_{p-j+1}}{p+1}< \frac{(n+1)^{p+1}}{p+1}.

i.e.,

(p+1)(1^p+2^p+3^p+\dots+n^p) <  (n+1)^{p+1}, \quad\quad p, n \in N^+.

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