Question

The following series do not satisfy the hypotheses of the alternating series test as stated. In each case, state which hypothesis is not satisfied. State whether the series converges absolutely.\(\sum_{n=1}^{\infty}(-1)^{n+1} \frac{\cos ^{2} n}{n}\)

Short answer

The series does not satisfy the decreasing term hypothesis and diverges absolutely.

Step-by-step solution

Identify the Series

The given series is \[\sum_{n=1}^{\infty} (-1)^{n+1} \frac{\cos^2 n}{n}\]This is an alternating series due to the factor \((-1)^{n+1}\).

Check if Terms Decrease

For the alternating series test, the sequence of terms \(b_n = \frac{\cos^2 n}{n}\) must be monotonically decreasing. Calculate:\[b_n = \frac{\cos^2 n}{n}\]However, because \(\cos^2 n\) is not a decreasing function of \(n\), the terms \(b_n\) do not necessarily decrease as \(n\) increases. Thus, this condition is not satisfied.

Check if the Limit of Terms is Zero

For alternating series, check whether \[\lim_{n \to \infty} \frac{\cos^2 n}{n} = 0\]As \(n\) approaches infinity, \(\frac{1}{n} \to 0\), but since \(\cos^2 n\) oscillates between 0 and 1, the overall limit approaches 0. So this condition is satisfied.

Evaluate Absolute Convergence

To check for absolute convergence, consider the series:\[\sum_{n=1}^{\infty} \left| \frac{\cos^2 n}{n} \right|\]Since \(\cos^2 n\) oscillates between 0 and 1:\[\frac{\cos^2 n}{n} \le \frac{1}{n}\]The series \(\sum_{n=1}^{\infty} \frac{1}{n}\) is the harmonic series, which is divergent. Hence, the series does not converge absolutely.

Key concepts

Absolute Convergence

When we talk about absolute convergence, it means examining if the series converges when we take the absolute value of its terms. This changes our series because we get rid of the alternating negative and positive signs. We decide absolute convergence by looking at:\[ \sum_{n=1}^{\infty} \left| \frac{\cos^2 n}{n} \right| \] This step is crucial when we think about convergence since absolute convergence is a stronger condition than regular convergence. If a series converges absolutely, it also converges normally (without absolute value). But, if it doesn't converge absolutely, it might still converge in the alternating form. For our series, the absolute series looks something like the harmonic series, given that:\[ \frac{\cos^2 n}{n} \le \frac{1}{n} \] Since the harmonic series \( \sum_{n=1}^{\infty} \frac{1}{n} \) is known to be divergent, we can conclude that our series does not converge absolutely.

Monotonically Decreasing Sequence

This concept checks if each term in a sequence gets smaller or stays the same as \(n\) increases. For an alternating series to pass a specific test (alternating series test), its sequence, without the sign change, should be monotonically decreasing. This means:- Each term should not be larger than the one before it.- In math terms, for \(b_n = \frac{\cos^2 n}{n}\), every \(b_n\) should satisfy \(b_{n+1} \le b_n\). However, because of the \(\cos^2 n\) part, which constantly changes its value as \(n\) increases, we can't promise this condition. Therefore, our series fails the test here. Understanding if a sequence decreases consistently helps us predict the behavior of a series as more and more terms are added. But since our terms don't decrease predictably, it casts doubt on the series converging using the alternating series test.

Harmonic Series

The harmonic series is a famous mathematical series. It takes the form \( \sum_{n=1}^{\infty} \frac{1}{n} \). What makes it intriguing is that, unlike some series that simply add up neat numbers, it grows in an unusual way.- The terms diminish very slowly. This slow decrease is what leads to its divergence, meaning it never settles down to a finite number even if you add a vast number of terms.Understanding the harmonic series is crucial because it acts as a benchmark for comparing other series. When determining if our original series converges absolutely, we compared it to the divergent harmonic series:- If our absolute series is smaller term by term than a divergent series like the harmonic series, it too must diverge. Despite the harmonic series having decreasing terms, it teaches us that not all decreasing series converge. In our problem, this comparison helps reinforce why \( \sum_{n=1}^{\infty} \left| \frac{\cos^2 n}{n} \right| \) does not converge.