Question

Use the Limit Comparison Test with the harmonic series to show that the series \(\sum a_{n}\) (where \(0

Short answer

Using the Limit Comparison Test with the harmonic series, it's shown that the given series \(\sum a_{n}\) diverges when the limit as n approaches infinity of \(na_{n}\) is a finite nonzero number.

Step-by-step solution

Define The Harmonic Series

The harmonic series is given by \(\sum^{\infty}_{n=1} \frac{1}{n}\). This series is known to be divergent, which will be helpful later

Apply The Limit Comparison Test

The Limit Comparison Test states that if the limit as n approaches infinity of \(\frac{a_{n}}{b_{n}}\) is a finite nonzero number c, then both series \(\sum_{n=1}^\infty a_{n}\) and \(\sum_{n=1}^\infty b_{n}\) either converge or diverge. In this case, we compare \(a_{n}\) with our harmonic series term \(\frac{1}{n}\) which forms our ratio.

Evaluate The Limit

The problem statement gives that the limit as n approaches infinity of \(n a_{n}\) is a finite nonzero number. Hence, by the Limit Comparison Test, our original series behaves in the similar way as the harmonic series. Since the known harmonic series is divergent, our given series also diverges.