Problem 12 - Single Variable Calculus

Question

Use the Comparison Test to determine whether the series is convergent or divergent. \(\sum_{n=1}^{\infty} \frac{1}{n^{n}}\)

Step-by-Step Solution

Answer

The given series is $\sum_{n=1}^{\infty} \frac{1}{n^n}$. We will use the Comparison Test to compare it with the convergent geometric series $\sum_{n=1}^{\infty} \frac{1}{2^n}$. We show that \(\frac{1}{n^n} \leq \frac{1}{2^n}\) for all n, which holds true as \(n^n \geq 2^n\). Since the comparison series converges, the given series $\sum_{n=1}^{\infty} \frac{1}{n^n}$ is also convergent by the Comparison Test.

Step 1: Write down the given series

The given series is \(\sum_{n=1}^{\infty} \frac{1}{n^n}\).

Step 2: Write down the comparison series

We will compare the given series with the geometric series \(\sum_{n=1}^{\infty} \frac{1}{2^n}\), which we know converges since its ratio \(\frac{1}{2} < 1\).

Step 3: Show the inequality between the two series

We want to show that \(\frac{1}{n^n} \leq \frac{1}{2^n}\) for all n. Rearrange the inequality to get: \[n^n \geq 2^n\] Since n is an integer greater than or equal to 1, this inequality will always hold true. For example: - If \(n = 1\), then \(1^1 \geq 2^1\), which is true. - For \(n \geq 2\), we see that increasing the base of the exponent from 2 to n will always produce a larger number. So, the inequality will hold for all \(n \geq 2\).

Step 4: Apply the Comparison Test

Since we have proven that \(\frac{1}{n^n} \leq \frac{1}{2^n}\), and we know that the comparison series \(\sum_{n=1}^{\infty} \frac{1}{2^n}\) converges, by the Comparison Test, we can conclude that the given series \(\sum_{n=1}^{\infty} \frac{1}{n^n}\) is convergent as well.

Key Concepts

Geometric Series

A geometric series is a series with a constant ratio between successive terms. In its simplest form, a geometric series can be expressed as:

• \( a + ar + ar^2 + ar^3 + \ldots \)

Here, \(a\) is the first term, and \(r\) is the common ratio. If the absolute value of the common ratio \(|r|\) is less than 1, the series will converge on a finite value. On the other hand, if \(|r|\) is equal to or greater than 1, the series will diverge. A classic example is the series \( \sum_{n=0}^{\infty} r^n \) which converges to \( \frac{1}{1-r} \) when \(|r| < 1\). In this context, understanding geometric series helps when using the Comparison Test, especially for determining convergence in others, like the series given in the exercise.

Series Convergence

Convergence in series refers to the property where the series' terms approach a finite limit as the number of terms grows infinitely. For a series \( \sum_{n=1}^{\infty} a_n \) to converge, the sequence of its partial sums must tend to a finite limit. To check convergence, various tests can be applied, such as:

• Geometric Series Test: Used when the series terms form a geometric series, checking the common ratio \( r \).
• Comparison Test: Compares the series with another known convergent or divergent series.
• Ratio Test: Looks at the limit of \( \frac{a_{n+1}}{a_n} \) as \( n \) tends to infinity.

In our example, we used the Comparison Test. The given series \( \sum_{n=1}^{\infty} \frac{1}{n^n} \) was compared to the geometric series \( \sum_{n=1}^{\infty} \frac{1}{2^n} \), known to converge. Through this comparison, convergence of the original series was confirmed.

Inequality in Series

Inequalities are important in comparing series and establishing convergence or divergence. Using inequalities, one can bound a series by another that has a known behavior. In the exercise, we used the inequality:

• \( \frac{1}{n^n} \leq \frac{1}{2^n} \)

To establish this inequality, we showed that \( n^n \geq 2^n \) for all \( n \geq 1 \). This comparison helped prove that \( \sum_{n=1}^{\infty} \frac{1}{n^n} \) is bounded above by a convergent series. Applying an inequality ensures that our smaller series remains convergent just like the larger comparator series. Such tactics are central in applying the Comparison Test correctly.

Infinite Series

An infinite series is the sum of infinitely many terms expressed in the form \( \sum_{n=1}^{\infty} a_n \). The concept hinges on examining the limit of partial sums rather than each term independently.Infinite series might converge (reach a finite sum) or diverge (increase without bounds) depending on the terms involved. When assessing infinite series, consider tests for convergence such as:

• Comparison Test
• Integral Test
• Ratio Test
• Root Test

For the series \( \sum_{n=1}^{\infty} \frac{1}{n^n} \), we used the Comparison Test with a known geometric series \( \sum_{n=1}^{\infty} \frac{1}{2^n} \). This example shows how an infinite series can converge even if terms diminish rapidly as \( n \) increases. Understanding infinite series involves identifying the properties of their terms and choosing the right convergence tests.