Problem 69 - Essential Calculus

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 $$ \sum_{n=1}^{\infty} \frac{n}{\left(n^{2}+1\right)^{2}} $$

Step-by-Step Solution

Answer

Applying each test to the series, we found that the nth-Term Test was inconclusive, the series couldn't be directly tested with the Geometric, p-series, and Telescoping Series Tests. The Integral Test and Limit Comparison Test resulted in the series converging, while the Direct Comparison Test was inconclusive. Conclusion is that the series converges.

Step 1: Nth-Term Test

For the nth-term test, we evaluate the limit of the nth term as \(n\) approaches infinity. If the limit is not zero, the series diverges. In this case, \(\lim_{n \to \infty} \frac{n}{(n^{2}+1)^{2}} = 0\). So, the nth-Term Test is inconclusive.

Step 2: Geometric Series Test

A geometric series takes the form \(\sum_{n=0}^{\infty} ar^n\), where \(a\) and \(r\) are constants. The given series does not have this form, so it can't be directly tested with Geometric Series Test.

Step 3: p-Series Test

A p-series takes the form \(\sum_{n=1}^{\infty} \frac{1}{n^p}\). Here the series is not in this form, so it can't be directly tested with the p-Series Test.

Step 4: Telescoping Series Test

The series does not contain terms that can be written as the difference of two terms, so we can't apply telescoping series test here.

Step 5: Integral Test

First, let's find a corresponding function for our series, which is \(f(n) = \frac{n}{(n^{2}+1)^{2}}\). Then carry out the integral of this function from 1 to infinity, \(\int_{1}^{\infty} \frac{x}{(x^{2}+1)^{2}} dx\). Evaluate this integral using substitution method, let \( u = x^2 + 1\). The integral converges. Hence by integral test, the series converges.

Step 6: Direct Comparison Test

With Direct Comparison test, we compare the terms of the series with the terms of a known series. Since \( \frac{n}{(n^{2}+1)^{2}} < \frac{1}{n}\), and the series \(\sum_{n=1}^{\infty} \frac{1}{n}\) is a divergent p-series, the Direct Comparison test is inconclusive.

Step 7: Limit Comparison Test

We can compare the given series to \(\sum_{n=1}^{\infty} \frac{1}{n^2}\), which is a convergent p-series. The limit comparison test involves finding the limit as \( n \to \infty \) of \(\frac{a_n}{b_n} = \frac{\frac{n}{(n^{2}+1)^{2}}}{\frac{1}{n^2}}\). Simplifying, we find this limit is 1, which means that since our comparison series converges, so does our given series.