Question

We will determine whether the series ∑a n =∑ n=1/[infinity] 4√n 5 +01 is convergent or divergent using the Limit Comparison Test (note that the Comparison Test could be used as well). The given series has positive terms, which is a requirement for applying the Limit Comparison Test. First we must find an appropriate series ∑b n for comparison (this series must also have positive terms). The most reasonable choice is b n =

Answer 1

To determine whether the series converges or diverges using the Limit Comparison Test, we need to find an appropriate series for comparison. The most reasonable choice is 1/n. Next, we apply the Limit Comparison Test by taking the limit of the ratio of the original series to the chosen series. If the limit is finite and nonzero, then the series converge or diverge in the same way. In this case, the limit is infinity, so we cannot determine the convergence or divergence of the original series.

Step-by-step explanation:

To determine whether the series ∑an = ∑n=1/∞ 4√n 5 +01 is convergent or divergent using the Limit Comparison Test, we need to find an appropriate series ∑bn for comparison. The most reasonable choice is bn = 1/n, which also has positive terms.

Next, we apply the Limit Comparison Test. Take the limit as n approaches infinity of the ratio of an to bn. If the limit is finite and nonzero, then both series converge or diverge in the same way.

Let's calculate the limit.

limn→∞ (4√n 5 +01) / (1/n)

limn→∞ (4√n 5 +01) * n

limn→∞ (4n^(3/2) 5n + n)

We can rewrite this limit as limn→∞ (4n^(3/2))/n + 5n/n + n/n

limn→∞ 4√n + 5 + 1

Finally, as n approaches infinity, both 4√n and 5 + 1 approach infinity, so the limit is infinity. Since the limit is not finite and nonzero, we cannot make any conclusions about the convergence or divergence of the original series.