Problem 58 - A Course in Calculus and Real Analysis

Question

For \(k \in \mathbb{N}\), let \(a_{k} \in \mathbb{R}\) with \(a_{k} \neq 0 .\) Define \(\alpha:=\lim \sup _{k \rightarrow \infty}\left|a_{k+1}\right| /\left|a_{k}\right|\) and \(\beta:=\liminf _{k \rightarrow \infty}\left|a_{k+1}\right| /\left|a_{k}\right| .\) Show that if \(\alpha<1\), then \(\sum_{k=1}^{\infty} a_{k}\) is absolutely convergent and if \(\beta>1\), then \(\sum_{k=1}^{\infty} a_{k}\) is divergent.

Step-by-Step Solution

Answer

Using the Ratio Test, we have shown that if \(\alpha < 1\), where \(\alpha = \limsup_{k\rightarrow\infty} \left|\frac{a_{k+1}}{a_k}\right|\), then the series \(\sum_{k=1}^\infty a_k\) converges absolutely. Conversely, if \(\beta > 1\), where \(\beta = \liminf_{k\rightarrow\infty}\left|\frac{a_{k+1}}{a_k}\right|\), the series \(\sum_{k=1}^\infty a_k\) is divergent.

Step 1: Definition of the Ratio Test

The Ratio Test states that given a series \(\sum_{k=1}^{\infty} a_{k}\), we can analyze its convergence properties by considering the limit of the absolute ratio of consecutive terms, called \(L\): $$ L = \lim_{k\rightarrow\infty}\left|\frac{a_{k+1}}{a_k}\right| $$ If \(L<1\), the series is absolutely convergent; if \(L>1\), the series is divergent; if \(L=1\), the test is inconclusive.

Step 2: Introduce \(\limsup\) and \(\liminf\) definitions

We are tasked with analyzing the series \(\sum_{k=1}^{\infty} a_{k}\), with \(\alpha:=\limsup_{k\rightarrow\infty} \left|\frac{a_{k+1}}{a_k}\right|\) and \(\beta := \liminf_{k\rightarrow\infty}\left|\frac{a_{k+1}}{a_k}\right|\). Let's first show that if \(\alpha < 1\), the series is absolutely convergent.

Step 3: Case \(\alpha < 1\)

Let \(r < \alpha\), such that \(0 < r < \alpha < 1\). Since \(\alpha\) is the \(\limsup\) of the absolute ratios, this means that there exists some integer \(K\) such that for all \(k > K\), we have \(\left|\frac{a_{k+1}}{a_k}\right| < r\). Now, let's compare the series to a geometric series with ratio \(r\). We have: $$ \left|\frac{a_{k+1}}{a_k}\right| < r \;\;\text{for all}\;\; k > K \implies |a_{k+1}| < r|a_k| $$ Iterating this relationship, we find that for \(k > K\): $$ |a_k| < |a_{K+1}|r^{k-(K+1)} $$ The series of the right-hand side \(\sum_{k>K} |a_{K+1}|r^{k-(K+1)}\) is a convergent geometric series since \(0 < r < 1\). By the comparison test, the series \(\sum_{k>K} |a_k|\) also converges. This further implies that the series \(\sum_{k=1}^\infty a_k\) converges absolutely.

Step 4: Case \(\beta > 1\)

Let \(r > \beta\), such that \(\beta < r < 1\). Since \(\beta\) is the \(\liminf\) of the absolute ratios, this means that for infinitely many values of \(k\), we have \(\left|\frac{a_{k+1}}{a_k}\right| > r > 1\). This implies that the absolute value of the ratio \(|a_{k+1}| / |a_k|\) doesn't tend to zero for infinitely many values of \(k\). Therefore, the term \(a_{k+1}\) doesn't tend to zero as \(k\rightarrow\infty\). By the basic divergence test for series, if the terms of the series \(a_k\) don't tend to zero, then the series must be divergent. Thus, the series \(\sum_{k=1}^\infty a_k\) is divergent when \(\beta > 1\).

Key Concepts

limsup and liminf properties

When examining sequences and series, understanding the behavior of ratios of consecutive terms is crucial. This is where the concepts of

• limsup (limit superior)
• liminf (limit inferior)

come into play. The idea is to capture the limiting behavior of the maxima and minima of a sequence, respectively.

**Limsup**, denoted as \( \limsup \), represents the smallest upper bound that the sequence approaches. It's particularly useful when determining if a series converges. If \( \alpha = \limsup_{k\rightarrow\infty} \left| \frac{a_{k+1}}{a_k} \right| < 1 \), the terms in the sequence eventually get smaller, indicating convergence.

**Liminf**, denoted as \( \liminf \), helps in understanding the smallest lower bound that the sequence approaches. If \( \beta = \liminf_{k\rightarrow\infty} \left| \frac{a_{k+1}}{a_k} \right| > 1 \), this suggests that the terms in the sequence don't decay to zero and instead grow, indicating divergence.

Overall, limsup and liminf give a comprehensive snapshot of a sequence's behavior, helping to apply the Ratio Test more effectively.

geometric series

Understanding a geometric series is vital for comprehending how comparison tests work.

A geometric series is a series of the form\[\sum_{k=0}^{\infty} ar^k = a + ar + ar^2 + ar^3 + \cdots\]where \(a\) is the first term and \(r\) is the common ratio.

The convergence of a geometric series depends on the absolute value of the ratio \(r\). If \(|r| < 1\), the series converges to \( \frac{a}{1-r} \). If \(|r| \geq 1\), the series diverges.

In applying the Ratio Test, if the ratio of consecutive terms \(\left|\frac{a_{k+1}}{a_k}\right|\) approaches a value less than 1, we often compare it to a geometric series. By doing this, we're able to determine convergence by noting that the smaller ratios imply a behavior similar to that of a converging geometric series.

This connection provides a straightforward method to prove absolute convergence efficiently.

divergence test for series

The divergence test is a fundamental tool to quickly assess whether a series can converge.

Before delving deep into more complex tests, one should always check if the terms of the series, \(a_k\), approach zero as \(k\rightarrow\infty\). The divergence test states:

• If \(\lim_{k\rightarrow\infty} a_k eq 0\), the series \(\sum_{k=1}^{\infty} a_k\) is divergent.
• If \(\lim_{k\rightarrow\infty} a_k = 0\), the test is inconclusive and further analysis is needed.

In the context of liminf, when \(\beta > 1\), the ratios of consecutive terms imply that the terms of the series do not diminish to zero. This violation of the necessary condition for convergence means that the series must diverge.

The divergence test provides a direct and immediate conclusion in scenarios where convergence cannot possibly occur, saving time and effort in analysis.