Show how the hyperharmonic series of order p, ∑(1/np) (p ∈ E1), converges if p > 1. Hence if p ≤ 1, then ∑_(n=1)^∞ (1/np) ≥ ∑_(n=1)^∞ (1/n) = +∞. or if p > 1, then ∑_(n=1)^∞ (1/np) = 1 + (1/2p + 1/3p) + (1/4p +...+1/7p) + (1/8p +...+1/15p)+...                                 ≤ 1 + (1/2p + 1/2p) + (1/4p +...+1/4p) + (1/8p +...+1/8p)+...                                 = ∑_(n=0)^∞ (2p-1)n, is a convergent geometric series.

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