Question

Determine whether the following series converge \((\theta\) and \(p\) are positive real numbers): (a) \(\sum_{n=1}^{\infty} \frac{2 \sin n \theta}{n(n+1)}\), (b) \(\sum_{n=1}^{\infty} \frac{2}{n^{2}}\), (c) \(\sum_{n=1}^{\infty} \frac{1}{2 n^{1 / 2}}\), (d) \(\sum_{n=2}^{\infty} \frac{(-1)^{n}\left(n^{2}+1\right)^{1 / 2}}{n \ln n}\), (e) \(\sum_{n=1}^{\infty} \frac{n^{p}}{n !}\).

Short answer

(a) Converges, (b) Converges, (c) Diverges, (d) Diverges, (e) Converges

Step-by-step solution

- Series (a) Analysis

To determine if \(\binom{n=1}{\to\text{∞}} \frac{2 \text{ sin } n \theta}{n(n+1)}\) converges, use the fact that \(\frac{\text{ sin } n \theta}{n(n+1)} \text{ is bounded and similar to } \frac{1}{n^2}.\) Compare it with a convergent p-series, \(\binom{k=1}{\to\text{∞}} \frac{1}{k^2}.\)

- Series (b) Analysis

Evaluate \(\binom{n=1}{\to\text{∞}} \frac{2}{n^2}\). This series is a p-series with p = 2 (greater than 1), which is known to converge.

- Series (c) Analysis

Evaluate \(\binom{n=1}{\to\text{∞}} \frac{1}{2n^{1/2}} \). This series is a p-series where p = 1/2 (less than 1), which is known to diverge.

- Series (d) Analysis

For \(\binom{n=2}{\text{to} \text{∞}} \frac{(-1)^n \big(n^2+1\big)^{1/2}}{n \text{ ln } n}\), consider the alternating series test (Leibniz criterion), but also analyze the behavior of \( \frac{\big(n^2+1\big)^{1/2}}{n \text{ ln } n}\). This term behaves similarly to \(\frac{1}{\text{ ln } n}\), which does not vanish as n tends to infinity, indicating divergence.

- Series (e) Analysis

For \(\binom{n=1}{\to\text{∞}} \frac{n^p}{n!}\) where p is positive, use the ratio test. \(a_n = \frac{n^p}{n!}\). Calculate \(\text{ lim }_{n \text{→∞}} \frac{a_{n+1}}{a_n} \text{ = lim }_{n \text{→∞}} \frac{(n+1)^p}{(n+1)!}\text{ }\frac{n!}{n^p} = \text{lim }_{n \text{→∞}} \frac{(n+1)^p}{n \text{ (n+1)}} = 0.\) Thus, the series converges.

Key concepts

p-series

A p-series is a series of the form \( \sum_{n=1}^{\text{∞}} \frac{1}{n^p} \), where p is a positive real number. The convergence or divergence of a p-series depends entirely on the value of p. Specifically: \

• When p > 1, the series converges.
• When p ≤ 1, the series diverges.

Note that p-series are essential in many problems involving series convergence, as they serve as a comparison tool to determine the behavior of other series. For example, consider the series \( \sum_{n=1}^{\text{∞}} \frac{2}{n^2} \). This series is a p-series with p = 2, which is greater than 1, so it converges. Similarly, \( \sum_{n=1}^{\text{∞}} \frac{1}{2n^{1/2}} \) has p = 1/2, which is less than or equal to 1, thus it diverges.

ratio test

The ratio test is a method used to determine the convergence or divergence of a series. To apply the ratio test, consider a series \( \sum_{n=1}^{\text{∞}} a_n \) and compute the limit: \[ L = \text{ lim }_{n \text{→∞}} \left| \frac{a_{n+1}}{a_n} \right| \. \] The behavior of the series is then determined by the value of L:

• If L < 1, the series converges absolutely.
• If L > 1, or if the limit does not exist, the series diverges.
• If L = 1, the test is inconclusive.

For example, in the series \( \sum_{n=1}=^{\text{∞}} \frac{n^p}{n!} \), applying the ratio test: \ \begin{align*} L & = \text{lim }_{n \text{→∞}} \left| \frac{a_{n+1}}{a_n} \right| \ &= \text{ lim }_{n \text{→∞}} \frac{(n+1)^p}{(n+1)!} \cdot \frac{n!}{n^p} \ &= \text{lim }_{n \text{→∞}} \frac{(n+1)^p}{n(n+1)} \ &= 0 \. \end{align*} Since 0 < 1, the series converges absolutely according to the ratio test.

alternating series test

The alternating series test, also known as the Leibniz criterion, is used to determine the convergence of an alternating series. An alternating series is any series of the form: \( \sum_{n=1}^{\text{∞}} a_n (-1)^n \). To apply the alternating series test, check the following conditions:

• \