Problem 71

Question

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

Step-by-Step Solution

Verified
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.
1Step 1: 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
2Step 2: 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.
3Step 3: 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.