Discussion
With n being a positive integer, the series \(\sum_{n=1}^{\infty} \frac{1}{n^{p}}\), for \(p \gt 1\) is
Answer: A
The series is convergent.
\(\sum 1/n^{p}\) is the p-series. By the integral test it behaves like \(\int_{1}^{\infty} x^{-p}\,dx\), which is finite exactly when \(p \gt 1\), so the series converges for \(p \gt 1\) and diverges for \(p \le 1\) (the harmonic series is the case \(p = 1\)).
It cannot be divergent here for that reason. 'Oscillatory' describes a series whose partial sums swing without settling, which needs terms of changing sign; every term here is positive, so the partial sums only increase. 'Asymptotic' is a property of expansions, not a convergence class of a series of positive terms.
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