Question

Test for convergence or divergence, using each test at least once. Identify which test was used. (a) \(n\) th-Term Test (b) Geometric Series Test (c) \(p\) -Series Test (d) Telescoping Series Test (e) Integral Test (f) Direct Comparison Test (g) Limit Comparison Test $$ \sum_{n=1}^{\infty} \frac{\sqrt{n}}{n} $$

Short answer

The series \(\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}\) diverges, as determined by the p-series test, direct comparison test, and limit comparison test. There was no need to use other tests for this series.

Step-by-step solution

Simplify the Series

First, simplify the series so it is easier to work with. The series \(\sum_{n=1}^{\infty} \frac{\sqrt{n}}{n}\) simplifies to \(\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}\)

Apply the p-Series Test

Now apply the p-series test. A p-series is a series of the form \(\sum_{n=1}^{\infty} \frac{1}{n^p}\). Comparing this to our series, we can see that \(p = \frac{1}{2}\). Since \(p \le 1\), we can say that the series is divergent according to the p-series test.

Apply the Direct Comparison Test

The Direct Comparison Test can also be used to confirm the result. Take the series \(\sum_{n=1}^{\infty} \frac{1}{n}\), which is known to diverge. Since \(\frac{1}{\sqrt{n}} \ge \frac{1}{n}\) for all \(n \ge 1\), the result from the Direct Comparison Test also indicates divergence.

Apply the Limit Comparison Test

Next, apply the Limit Comparison Test with the series \(\sum_{n=1}^{\infty} \frac{1}{n}\). The limit \(\lim_{n \rightarrow \infty} \frac{\frac{1}{\sqrt{n}}}{\frac{1}{n}}\) simplifies to \( \lim_{n \rightarrow \infty} \sqrt{n}\), which is infinite. Thus, according to the Limit Comparison Test, since our series behaves similarly to a known divergent series, our series must also diverge.

Conclusion

Since multiple tests have confirmed that the series diverges, we can be confident in this result. It is not necessary to apply every test to this series to determine convergence or divergence. The result from each test was consistent, giving strong evidence for divergence.