Question

Determine convergence or divergence. (a) \(\sum \frac{n^{n}\left(2+(-1)^{n}\right)}{2^{n}}\) (b) \(\sum\left(\frac{1+\sin 3 n \theta}{3}\right)^{n}\) (c) \(\sum(n+1)\left(\frac{1+\sin (n \pi / 6)}{3}\right)^{n}\) (d) \(\sum\left(1-\frac{1}{n}\right)^{n^{2}}\)

Short answer

Based on the analysis and tests performed, we can conclude the following about the convergence or divergence of each series: (a) \(\sum \frac{n^{n}\left(2+(-1)^{n}\right)}{2^{n}}\) is divergent. (b) \(\sum\left(\frac{1+\sin 3 n \theta}{3}\right)^{n}\) is convergent. (c) \(\sum(n+1)\left(\frac{1+\sin (n \pi / 6)}{3}\right)^{n}\) is divergent. (d) For \(\sum\left(1-\frac{1}{n}\right)^{n^{2}}\), we are unable to determine convergence or divergence with the information given.

Step-by-step solution

Analyze the given series

Notice that the series has a term \((-1)^n\), suggesting that there might be an alternating pattern in the series. Let's verify that by finding the first few terms of the series for n=1, 2, 3 and 4: When n=1, the term is \(\frac{1(3)}{2}=\frac{3}{2}.\) When n=2, the term is \(\frac{2^{2}(2-1)}{2^{2}}=\frac{2}{2}=1.\) When n=3, the term is \(\frac{3^{3}(2+1)}{2^{3}}=\frac{27(3)}{8}.\) When n=4, the term is \(\frac{4^{4}(2-1)}{2^{4}}=\frac{256}{16}=16.\) As we can see the series doesn't seem to be alternating because the terms are not decreasing in absolute value, nor is it consistently alternating between positive and negative terms. Thus, we should try another convergence test.

Apply Ratio Test

Applying the Ratio Test, we have to find the limit $$\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| =\lim_{n\to\infty}\left|\frac{(n+1)^{(n+1)}\left( 2+(-1)^{(n+1)}\right)}{2^{n+1}}\cdot\frac{2^n}{n^n\left(2+(-1)^n\right)} \right|.$$ Using algebraic manipulations, we get $$\lim_{n\to\infty}\left|\frac{n^{n}\left(2+(-1)^n\right)}{(n+1)^n}\cdot\frac{(n+1)(2+(-1)^{n+1})}{2}\right|.$$ Now, use the limit laws and the limit of the product is the product of the limits: $$\lim_{n\to\infty}\frac{(n+1)(2+(-1)^{n+1})}{2}\lim_{n\to\infty}\left(\frac{n^{n}}{(n+1)^{n}}\cdot\frac{2+(-1)^n}{2+(-1)^{n+1}}\right).$$ The first limit doesn't exist because its value oscillates between different values. Thus, using the Ratio Test, the overall limit doesn't exist, and we conclude that the series is divergent. (b) Determine convergence or divergence: \(\sum\left(\frac{1+\sin 3 n \theta}{3}\right)^{n}\)

Apply Root Test

Applying the Root Test, we have to find the limit $$\lim_{n\to\infty}\sqrt[n]{\left|\frac{1+\sin 3 n \theta}{3}\right|^n}.$$ By definition of the n-th root, we get $$\lim_{n\to\infty}\left|\frac{1+\sin 3 n \theta}{3}\right|.$$ Since \(0\leq\left|\sin(3n\theta)\right|\leq1\), we know that \(0\leq\left|\frac{1+\sin 3 n \theta}{3}\right|\leq1\). Thus, by the Squeeze Theorem, the limit will be a value between 0 and 1. From the Root Test, we know that if the limit is less than 1, the series is absolutely convergent. Hence, the given series is convergent. (c) Determine convergence or divergence: \(\sum(n+1)\left(\frac{1+\sin (n \pi / 6)}{3}\right)^{n}\)

Apply Comparison Test

Notice that each term has the form \((n+1)y^n\) where \(0 \leq y = \left(\frac{1+\sin(n\pi/6)}{3}\right) \leq \frac{4}{3}\). Hence, we can use the Comparison Test to analyze the given series. We have: $$0\leq(n+1)\left(\frac{1+\sin (n \pi / 6)}{3}\right)^{n}\leq(n+1)\left(\frac{4}{3}\right)^n$$ Now, let's consider the series \(\sum(n+1)\left(\frac{4}{3}\right)^n\).

Apply Ratio Test to the comparison series

Applying the Ratio Test to the comparison series, we have to find the limit $$\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| =\lim_{n\to\infty}\left|\frac{(n+2)\left(\frac{4}{3}\right)^{n+1}}{(n+1)\left(\frac{4}{3}\right)^n}\right|=\lim_{n\to\infty}\left|\frac{4(n+2)}{3(n+1)}\right|.$$ The limit is $$\lim_{n\to\infty}\left|\frac{4(n+2)}{3(n+1)}\right|= \frac{4}{3}>1.$$ Hence, the comparison series \(\sum(n+1)\left(\frac{4}{3}\right)^n\) is divergent. Therefore, the original series \(\sum(n+1)\left(\frac{1+\sin (n \pi / 6)}{3}\right)^{n}\) is also divergent by the Comparison Test. (d) Determine convergence or divergence: \(\sum\left(1-\frac{1}{n}\right)^{n^{2}}\)

Apply the Root Test

Applying the Root Test, we have to find the limit $$\lim_{n\to\infty}\sqrt[n^2]{\left|\left(1-\frac{1}{n}\right)^{n^2}\right|}.$$ By definition of the n^2-th root, we get $$\lim_{n\to\infty}\left|1-\frac{1}{n}\right|.$$ The limit is $$\lim_{n\to\infty}\left|1-\frac{1}{n}\right|=1.$$ As the limit is equal to 1, the Root Test is inconclusive. We need to use another test to determine the convergence or divergence of the series.

Apply the Divergence Test

Applying the Divergence Test, we have to find the limit $$\lim_{n\to\infty}\left(1-\frac{1}{n}\right)^{n^{2}}.$$ Using the definition of limit and the properties of exponentials, we get $$\lim_{n\to\infty}\left(e^{-\frac{1}{n}}\right)^{n^{2}}=e^{\lim_{n\to\infty}-n}.$$ The limit is $$e^{\lim_{n\to\infty}-n}=0.$$ As the limit of the series is 0, the Divergence Test is also inconclusive. Since we cannot use the Comparison Test, Limit Comparison Test, or Root Test here, we cannot determine the convergence or divergence of the series with the information given.

Key concepts

Ratio Test

The Ratio Test is a powerful tool to determine the convergence or divergence of an infinite series. It involves evaluating the limit of the absolute value of the ratio of consecutive terms. Given a series \( \sum a_n \), you consider the expression \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \).

• If the result of this limit is less than 1, the series is absolutely convergent.
• If the limit is greater than 1, the series is divergent.
• If the limit equals 1, the test is inconclusive.

The Ratio Test is particularly useful when dealing with series that involve factorials or exponential terms, as it simplifies the algebra significantly. While it's a very handy tool, remember that it won't always give you a definitive answer, especially if the limit equals 1. In these cases, other tests might be necessary to fully understand the behavior of the series.

Root Test

The Root Test, similar to the Ratio Test, helps in assessing the convergence of a series. This test calculates the n-th root of the absolute value of the terms, specifically the limit \( \lim_{n \to \infty} \sqrt[n]{|a_n|} \).

• If this limit is less than 1, the series converges absolutely.
• If the limit is greater than 1, the series diverges.
• As with the Ratio Test, if the limit is exactly 1, the test provides no information (inconclusive).

The Root Test is especially effective for series where terms involve powers of n. Terms that rise exponentially are particularly suited for this analysis since the test works directly with roots. However, like all convergence tests, it's important to choose the correct test based on the nature of the series you are examining.

Alternating Series

When a series flips sign between consecutive terms, it is known as an Alternating Series. Such series are often of the form \( \sum (-1)^n a_n \) or \( \sum (-1)^{n+1} a_n \). The Alternating Series Test is a crucial method to determine the convergence of these series, stating that:

• The terms \( a_n \) must be positive.
• The terms \( a_n \) should be decreasing, meaning \( a_{n+1} \leq a_n \).
• The limit of \( a_n \) as \( n \rightarrow \infty \) must be 0.

If all these conditions are met, the series converges. Alternating Series can also be conditionally or absolutely convergent. Absolute convergence implies that the series \( \sum |a_n| \) also converges, while conditional convergence occurs when \( \sum a_n \) converges, but \( \sum |a_n| \) does not. It’s this flip of sign in every term that adds a stability helping the sequence approach a limit.

Comparison Test

The Comparison Test allows you to determine the convergence of a series by comparing it to another series whose convergence behavior is known. Suppose you have two series: \( a_n \) and \( b_n \). To apply the Comparison Test:

• If \( 0 \leq a_n \leq b_n \) and \( \sum b_n \) converges, then \( \sum a_n \) also converges.
• If \( 0 \leq b_n \leq a_n \) and \( \sum b_n \) diverges, then \( \sum a_n \) also diverges.

This test is particularly helpful when you can recognize that the series in question can be bounded by a simpler series. For instance, it simplifies determining the convergence of complicated terms by comparing them to a geometric or p-series. However, if it’s challenging to find a suitable comparison series, the Limit Comparison Test provides an alternative. It involves evaluating the limit of \( \frac{a_n}{b_n} \) as \( n \to \infty \), offering more flexibility in many scenarios.