Problem 43 - Essential Calculus Early Transcendentals

Question

For which of the following series is the Ratio Test inconclusive (that is, it fails to give a definite answer)? (a) $$\sum_{n=1}^{\infty} \frac{1}{n^{3}}$$ (b) $$\sum_{n=1}^{\infty} \frac{n}{2^{n}}$$ (c) $$\sum_{n=1}^{\infty} \frac{(-3)^{n-1}}{\sqrt{n}}$$ (d) $$\sum_{n=1}^{\infty} \frac{\sqrt{n}}{1+n^{2}}$$

Step-by-Step Solution

Answer

Series (a) and (d) are inconclusive with the Ratio Test.

Step 1: Formulate the Ratio Test

The Ratio Test involves taking the limit \( L = \lim_{{n \to \infty}} \frac{{a_{n+1}}}{{a_n}} \) where \( a_n \) is the general term of the series. Based on the value of \( L \), we can conclude whether a series converges: if \( L < 1 \), the series converges; if \( L > 1 \), the series diverges; and if \( L = 1 \), the test is inconclusive.

Step 2: Apply Ratio Test to Series (a)

Consider the series \( \sum_{n=1}^{\infty} \frac{1}{n^3} \). Here, \( a_n = \frac{1}{n^3} \). The ratio \( \frac{a_{n+1}}{a_n} = \frac{1/(n+1)^3}{1/n^3} = \left(\frac{n}{n+1}\right)^3 \). Taking the limit as \( n \to \infty \), we get \( \lim_{n \to \infty} \left(\frac{n}{n+1}\right)^3 = 1 \).

Step 3: Apply Ratio Test to Series (b)

Consider the series \( \sum_{n=1}^{\infty} \frac{n}{2^n} \). Here, \( a_n = \frac{n}{2^n} \). The ratio \( \frac{a_{n+1}}{a_n} = \frac{(n+1)/2^{n+1}}{n/2^n} = \frac{n+1}{2n} \). Taking the limit as \( n \to \infty \), we find \( \lim_{n \to \infty} \frac{n+1}{2n} = \frac{1}{2} \ < 1 \).

Step 4: Apply Ratio Test to Series (c)

Consider the series \( \sum_{n=1}^{\infty} \frac{(-3)^{n-1}}{\sqrt{n}} \). Here, \( a_n = \frac{(-3)^{n-1}}{\sqrt{n}} \). The ratio \( \frac{a_{n+1}}{a_n} = \frac{(-3)^n/\sqrt{n+1}}{(-3)^{n-1}/\sqrt{n}} = -\frac{3\sqrt{n}}{\sqrt{n+1}} \). Taking the limit as \( n \to \infty \), we have \( \lim_{n \to \infty} -\frac{3\sqrt{n}}{\sqrt{n+1}} = -3 \), whose magnitude is \( 3 \ > 1 \).

Step 5: Apply Ratio Test to Series (d)

Consider the series \( \sum_{n=1}^{\infty} \frac{\sqrt{n}}{1+n^2} \). Here, \( a_n = \frac{\sqrt{n}}{1+n^2} \). The ratio \( \frac{a_{n+1}}{a_n} = \frac{\sqrt{n+1}/(1+(n+1)^2)}{\sqrt{n}/(1+n^2)} \). Simplifying gives \( \frac{\sqrt{n+1}(1+n^2)}{\sqrt{n}(1+(n+1)^2)} \). Taking the limit as \( n \to \infty \), this necessarily involves manipulating terms, eventually leading to \( 1 \).

Step 6: Determine the Inconclusive Scenario

From the computations, series (a) and (d) provide a ratio limit \( L = 1 \), which makes the Ratio Test inconclusive as required. Series (b) has \( L < 1 \) indicating convergence, and series (c) has \(|L| > 1\) which means it diverges.

Key Concepts

Convergent Series

A convergent series is one that approaches a specific value as you sum its terms one by one to infinity. Think of it as getting closer and closer to a number without ever actually reaching it, much like an endless marathon that you almost finish.
The terms in the series become very small as they extend towards infinity, making the overall sum finite.

• Mathematically, a series \( \sum_{n=1}^{\infty} a_n \) is convergent if the sequence of partial sums \( S_n = \sum_{k=1}^{n} a_k \) approaches a limit \( L \) as \( n \to \infty \).

For example, in the context of the Ratio Test, if the limit \( L < 1 \) after applying the test, the series is considered convergent.
In such cases, you can be certain about the series settling to a nice, well-defined value despite adding an infinite number of terms.

Divergent Series

On the flip side, a divergent series doesn't settle down to any particular value when its terms are added, even if you go to infinity.
Unlike convergent series, divergent series have terms that are too big to result in a finite sum.

• If after applying the Ratio Test the limit \( L > 1 \), it indicates divergence.
• The series keeps increasing or oscillates indefinitely, making it impossible to pin down a single number as its sum.

For instance, with the series \( \sum_{n=1}^{\infty} \frac{(-3)^{n-1}}{\sqrt{n}} \), the test yields an \( |L| > 1 \), confirming divergence.
Divergent series can never be equal to a finite, reasonable figure as the sum grows without bound or stabilizes erratically.

Limit of a Series

The limit of a series is like the destination you are trying to reach through endless journeying by summing up its terms.
For a series to have a limit, it means the series' total keeps approaching a certain finite number, no matter how many terms you add.

• This number, \( L \), exists only if the series is convergent, suggesting the whole sum tends towards this value as you add up all the pieces.
• If \( L \) is equal to 1 while using the Ratio Test, it becomes a puzzling scenario since the test doesn't conclude whether the series converges or diverges.

Therefore, having no definite limit is indicative of a series like \( \sum_{n=1}^{\infty} \frac{1}{n^3} \) or \( \sum_{n=1}^{\infty} \frac{\sqrt{n}}{1+n^2} \), where the Ratio Test shows \( L = 1 \), pointing out the uncertainty of a clear limit.
The series seems to hang in balance without getting a decisive answer.

Infinite Series

An infinite series involves an endless addition of terms without a hard stop.
Unlike finite series that have a clear beginning and end, infinite series extend indefinitely, challenging us to find if they actually converge or not.

• Infinite series can transform into piles of convergence or divergence, based on their term behavior.
• The Ratio Test helps analyze their limits, guiding us to conclusions about convergence when possible.

For example, an infinite series like \( \sum_{n=1}^{\infty} \frac{n}{2^n} \) goes on forever, yet the Ratio Test clarifies it converges since \( L < 1 \).
Infinite series are crucial in mathematics for uncovering patterns and values hidden in limitless additions, sometimes resulting in familiar mathematical constants or behavior insights.