Question

Divergent case of harmonic p series (1/(n^p))

A) When p ≤ 1
B) When p > 1
C) When p = 1
D) When p approaches infinity

Answer 1

The divergent case of harmonic p series (1/(n^p)) occurs when p ≤ 1, including the special case when p = 1, known as the harmonic series, which also diverges. For p > 1, the series is convergent. When p approaches infinity, the series trivially converges since all terms other than the first are zero.

Step-by-step explanation:

The divergent case of harmonic p series (1/(n^p)) occurs in different scenarios depending on the value of p. For values of p ≤ 1, the series diverges. Specifically, case A applies when p ≤ 1, as the terms of the series do not decrease rapidly enough to result in a finite sum. Case B, when p > 1, actually represents the convergent case, where the series does sum to a limited value. Case C, where p = 1, is a special case of A and is known as the harmonic series, which famously diverges. Lastly, Case D, when p approaches infinity, is not relevant to the divergence or convergence of the series because it is a situation where the series would consist of only zeros after the first term, and thus is trivially convergent.