Question

State whether each of the following series converges absolutely, conditionally, or not at all\(\sum_{n=1}^{\infty}(-1)^{n+1}\left(\frac{1}{n}-\frac{1}{n+1}\right)\)

Short answer

The series converges conditionally.

Step-by-step solution

Rewrite the Series

Start by expressing the general term of the series: \((-1)^{n+1}\left(\frac{1}{n}-\frac{1}{n+1}\right)\). Observe that \((-1)^{n+1}\left(\frac{1}{n}-\frac{1}{n+1}\right)\) is an alternating series where each term follows the pattern of the difference of two fractions.

Simplify the General Term

The general term can be rewritten as \(a_n = \left(\frac{1}{n} - \frac{1}{n+1}\right) = \frac{1}{n(n+1)}\).This indicates that the series can be considered as:\(a_n = \frac{1}{n} - \frac{1}{n+1}\).Notice here that this part shows telescoping properties.

Check for Absolute Convergence

To check for absolute convergence, consider the absolute value of \(a_n\) without the alternating sign factor:\(\sum_{n=1}^{\infty} \left| \frac{1}{n(n+1)} \right|\).Since \(a_n = \frac{1}{n(n+1)}\) behaves like a telescoping series, we compare it with \(\sum_{n=1}^{\infty} \left(\frac{1}{n} - \frac{1}{n+1} \right)\) which simplifies and is known to be convergent. Therefore, it doesn't absolutely converge.

Check for Conditional Convergence

Check if the original series converges using the alternating series test. Note:\((-1)^{n+1}\) introduces alternating behavior.For convergence of an alternating series:1. \(b_n = \frac{1}{n(n+1)}\) should decrease and approach 0 as \(n\rightarrow\infty\).Both conditions are satisfied making the series conditionally converging.

Key concepts

Alternating Series

An alternating series is a series where the signs of its terms alternate between positive and negative. The series \[-(-1)^{n+1}\left(\frac{1}{n}-\frac{1}{n+1}\right)\]is an example of an alternating series because it includes the factor \((-1)^{n+1}\),which alternates the signs of the terms depending on whether \(n\) is odd or even. Alternating series can exhibit special properties that affect their convergence.

To determine if an alternating series converges, we use the Alternating Series Test. This test states that if the absolute values of the terms decrease steadily to zero, that is, \(b_{n+1} \leq b_n\)for all \(n\), and\(b_n \rightarrow 0\), the series converges.

In practical terms, alternating series often converge even if the series formed by the absolute values of the terms doesn't. Thus, identifying an alternating pattern in a series can be a helpful first step in determining its convergence properties.

Absolute Convergence

Absolute convergence refers to a series that converges even when all of its terms are replaced by their absolute values. For the series \[\sum_{n=1}^{\infty} (-1)^{n+1}\left(\frac{1}{n}-\frac{1}{n+1}\right)\],to check for absolute convergence, we consider the series formed by taking the absolute value of each term:\[\sum_{n=1}^{\infty} \left|\frac{1}{n} - \frac{1}{n+1}\right|.\]

In this case, the series becomes \[\sum_{n=1}^{\infty} \frac{1}{n(n+1)}.\]This series does not satisfy the requirements for absolute convergence because it behaves like a geometric series that doesn't converge when considering only its absolute values.

Understanding absolute convergence is crucial because it implies stability under changing terms. If a series converges absolutely, reordering its terms does not affect the sum. Unlike conditional convergence, this makes the series more stable and predictable.

Conditional Convergence

Conditional convergence occurs in series that converge only due to the alternating behavior of their terms, rather than their absolute values converging. In the case of \[\sum_{n=1}^{\infty} (-1)^{n+1}\left(\frac{1}{n}-\frac{1}{n+1}\right),\]we can apply the Alternating Series Test. The test verifies that the absolute values of terms decrease steadily and approach zero:

• The sequence \(\frac{1}{n(n+1)}\) decreases as \(n\) increases.
• As \(n\) approaches infinity, \(\frac{1}{n(n+1)} \to 0\).

Since these conditions are satisfied, the series converges conditionally.

Conditional convergence highlights the delicate balance that exists in certain series. Unlike absolute convergence, any reordering of terms in a conditionally convergent series can potentially change its sum. This property underscores the importance of understanding the series' structure before determining convergence.

Telescoping Series

A telescoping series derives its name from how its terms cancel out like a telescope collapsing. This series type often simplifies the process of finding its sum by reducing the expression within the series itself. The given series, with general term \[\frac{1}{n} - \frac{1}{n+1},\]exhibits this telescoping behavior.

When you write out the first few terms, you'll notice that most terms cancel out with subsequent ones:\[\left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \cdots\]Only the initial term \(\frac{1}{1}\) survives because each term cancels with the next term in the series.

The magic of telescoping is its ability to transform an infinite series into a finite one for evaluation. Once telescoped, what's left is often straightforward to evaluate, making these series particularly interesting and useful in mathematical analysis.