Question

Show that the series converges absolutely. (a) \(\sum(-1)^{n} \frac{1}{n(\log n)^{2}}\) (b) \(\sum \frac{\sin n \theta}{2^{n}}\) (c) \(\sum(-1)^{n} \frac{1}{\sqrt{n}} \sin \frac{\pi}{n}\) (d) \(\sum \frac{\cos n \theta}{\sqrt{n^{3}-1}}\)

Short answer

a) \(\sum(-1)^{n} \frac{1}{n(\log n)^{2}}\) b) \(\sum \frac{\sin n \theta}{2^{n}}\) c) \(\sum(-1)^{n} \frac{1}{\sqrt{n}} \sin \frac{\pi}{n}\) d) \(\sum \frac{\cos n \theta}{\sqrt{n^{3}-1}}\) Answer: a), b), and d) converge absolutely, while c) converges conditionally.

Step-by-step solution

Take the absolute value of the series

To analyze absolute convergence, we will first rewrite the series with the absolute value: \(\sum \left|\frac{(-1)^{n}}{n(\log n)^{2}}\right| = \sum \frac{1}{n (\log n)^{2}}\)

Apply Comparison Test

Since \(\frac{1}{n(\log n)^{2}}\) is a positive term, we can apply the Comparison Test. We will compare this term to \(\frac{1}{n \log n}\).

Evaluate the limit of the ratio

Evaluate the limit of the ratio between these two terms: \(\lim_{n \to \infty} \frac{\frac{1}{n (\log n)^2}}{\frac{1}{n \log n}} = \lim_{n \to \infty} \frac{\log n}{(\log n)^2} = \lim_{n \to \infty} \frac{1}{\log n}\).

Determine the convergence

Since \(\lim_{n \to \infty} \frac{1}{\log n}\) is a p-series with p > 1, it converges. Therefore, the original series \(\sum \frac{1}{n (\log n)^{2}}\) converges by the Limit Comparison Test. Hence, the given series converges absolutely. (b) \(\sum \frac{\sin n \theta}{2^{n}}\)

Take the absolute value of the series

To analyze absolute convergence, we will first rewrite the series with the absolute value: \(\sum \left|\frac{\sin n\theta}{2^n}\right| \leq \sum \frac{1}{2^n}\)

Apply Direct Comparison Test

Since the terms of the series \(\sum \frac{1}{2^n}\) are positive and the inequality holds true, we can apply the Direct Comparison Test.

Determine the convergence

The series \(\sum \frac{1}{2^n}\) is a geometric series with a ratio 1/2 which is less than 1. Therefore, it converges. Since the inequality holds true and the comparison series converges, we can conclude that the given series converges absolutely. (c) \(\sum(-1)^{n} \frac{1}{\sqrt{n}} \sin \frac{\pi}{n}\)

Take the absolute value of the series

To analyze absolute convergence, we will first rewrite the series with the absolute value: \(\sum \left|\frac{(-1)^{n}}{\sqrt{n}} \sin \frac{\pi}{n}\right| \leq \sum \frac{1}{\sqrt{n}}\)

Apply Direct Comparison Test

Since the terms of the series \(\sum \frac{1}{\sqrt{n}}\) are positive and the inequality holds true, we can apply the Direct Comparison Test.

Determine the convergence

The series \(\sum \frac{1}{\sqrt{n}}\) is a p-series with p = 1/2, which is less than 1, so it diverges. However, by the Alternating Series Test, the series \(\sum(-1)^{n} \frac{1}{\sqrt{n}} \sin \frac{\pi}{n}\) converges conditionally. Since the comparison series in Step 2 (which is greater than the original series) diverges, we have shown that our original series does not converge absolutely. (d) \(\sum \frac{\cos n \theta}{\sqrt{n^{3}-1}}\)

Take the absolute value of the series

To analyze absolute convergence, we will first rewrite the series with the absolute value: \(\sum \left|\frac{\cos n\theta}{\sqrt{n^3-1}}\right| \leq \sum \frac{1}{\sqrt{n^3-1}}\)

Apply Comparison Test

Since the terms of the series \(\sum \frac{1}{\sqrt{n^3-1}}\) are positive, we can apply the Comparison Test.

Evaluate the limit of the ratio

Evaluate the limit of the ratio between these two terms: \(\lim_{n \to \infty} \frac{\frac{1}{\sqrt{n^3-1}}}{\frac{1}{n^{3/2}}} = \lim_{n \to \infty} \frac{n^{3/2}}{\sqrt{n^3-1}} = 1\).

Determine the convergence

Since \(\lim_{n \to \infty} \frac{1}{n^{3/2}}\) is a p-series with p = 3/2 > 1, it converges. Therefore, the original series \(\sum \frac{1}{\sqrt{n^{3}-1}}\) also converges by the Limit Comparison Test. Hence, the given series converges absolutely.

Key concepts

Comparison Test

The Comparison Test is a powerful tool in determining the convergence of infinite series, especially when the series' terms are all non-negative. It involves comparing the series in question to another series whose behavior is well-known. For example, if every term of our series is smaller than or equal to the corresponding term of a convergent series, then our series must also converge.

This approach works because, intuitively, if the 'benchmark' series manages to add up to a finite number despite having larger terms, our series, which is made of smaller terms, will also add up to some finite number. The inverse also holds - if every term of our series is larger than the corresponding term of a divergent series, then our series must also diverge, as it's 'weightier' than the already divergent series.

Direct Comparison Test

Very similar to the Comparison Test is the Direct Comparison Test, which is even more straightforward. It requires comparing the terms of two series directly, where one series acts as the 'control' or 'standard' for comparison. When the terms of the series we're testing are all less than the corresponding terms of a known convergent series, and greater than zero, our series will also converge.

Conversely, if the terms of our series are greater than those of a known divergent series, ours must diverge too. The Direct Comparison Test is particularly useful because it keeps things simple - if you can show the inequality holds true for all terms, you often don't need to calculate anything else.

Alternating Series Test

The Alternating Series Test is specially designed for series where the signs of the terms alternate between positive and negative. For a series to pass this test, and thereby prove convergence, the absolute values of the terms need to be decreasing and approach zero as 'n' goes to infinity.

This test works based on the principle that the alternating signs cause each term to partially cancel out the one before it. This 'cancellation effect' is sufficient for convergence but does not necessarily imply absolute convergence. When a series converges by the Alternating Series Test but fails to converge when all terms are replaced by their absolute values, it's considered conditionally convergent.

p-series

A p-series is defined as an infinite series of the form \(\sum \frac{1}{n^p}\), where 'p' is a constant. The convergence of a p-series depends entirely on the value of 'p'. If 'p' is greater than 1, the series converges, known as the p-test. However, if 'p' is 1 or less, the series diverges.

The p-series is a cornerstone of series convergence tests because of its simple exponent rule for convergence, making it an ideal candidate for comparison with other more complex series. If we can express or compare the terms of an unknown series to a p-series, we can often determine the convergence or divergence of the original series with relative ease.

geometric series

A geometric series is an infinite series of the form \(\sum ar^n\), where 'a' is the first term and 'r' is the common ratio between consecutive terms. This series converges if the absolute value of 'r' is less than 1 and diverges if it is equal to or greater than 1. This criterion is due to the multiplying effect of the ratio: if it's less than one, subsequent terms diminish over time, leading to a finite sum.

Geometric series are significant in convergence tests because they have a straightforward criterion for convergence and a well-understood sum when they do converge, which makes them an excellent benchmark for the Comparison and Direct Comparison Tests. Recognizing a geometric series or recasting a complex series into a geometric one can greatly simplify the problem of determining convergence.