Problem 21 - Thomas Calculus

Question

In Exercises \(17-46,\) use any method to determine if the series converges or diverges. Give reasons for your answer. $$ \sum_{n=1}^{\infty} \frac{n^{10}}{10^{n}} $$

Step-by-Step Solution

Answer

The series \( \sum_{n=1}^{\infty} \frac{n^{10}}{10^n} \) converges by the Ratio Test.

Step 1: Identify the Series Type

The given series is \( \sum_{n=1}^{\infty} \frac{n^{10}}{10^n} \). This is a series with terms of the form \( \frac{f(n)}{r^n} \), where \( f(n) = n^{10} \). It resembles a form suitable for the Ratio Test due to the exponential \( 10^n \) in the denominator.

Step 2: Apply the Ratio Test

The Ratio Test states that for a series \( \sum a_n \), if \( \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right| = L \) and \( L < 1 \), the series converges. Here, \( a_n = \frac{n^{10}}{10^n} \).Find \( a_{n+1} = \frac{(n+1)^{10}}{10^{n+1}} \) and compute the ratio:\[\left| \frac{a_{n+1}}{a_n} \right| = \frac{(n+1)^{10}}{10^{n+1}} \cdot \frac{10^n}{n^{10}} = \frac{(n+1)^{10}}{10 \cdot n^{10}}\]

Step 3: Simplify the Ratio

Simplify the ratio derived in Step 2:\[\frac{(n+1)^{10}}{10 \cdot n^{10}} = \frac{1}{10} \left( \frac{n+1}{n} \right)^{10} = \frac{1}{10} \left( 1 + \frac{1}{n} \right)^{10}\]

Step 4: Evaluate the Limit

Evaluate the limit:\[L = \lim_{n\to\infty} \frac{1}{10} \left( 1 + \frac{1}{n} \right)^{10} = \frac{1}{10} \times 1^{10} = \frac{1}{10}\]Since \( L = \frac{1}{10} < 1 \), by the Ratio Test, the series converges.

Key Concepts

Ratio Test

The Ratio Test is a powerful tool used to determine the convergence of a series. If you have a series like \( \sum a_n \), the Ratio Test suggests you examine the limit of the absolute value of the ratio of consecutive terms. If

• \( L = \lim_{n\to\infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 \), the series converges.
• \( L > 1 \) or the limit is infinite, the series diverges.
• \( L = 1 \), the test is inconclusive.

In our example, the series \( \sum_{n=1}^{\infty} \frac{n^{10}}{10^n} \) fits perfectly for the Ratio Test due to the exponential term \( 10^n \) in the denominator. We derived the limit \( L = \frac{1}{10} \), showing the series converges.

Infinite Series

An infinite series is a sum of an infinite sequence of terms. Each term is defined by a specific formula involving an index like \( n \). Series often take forms where terms shrink exponentially as \( n \) increases.

Infinite series can sum to a finite value, converge, or can grow indefinitely, diverge.
Convergence and divergence depend on how the terms behave as \( n \) approaches infinity. Math tools like the Ratio Test help us identify the behavior. In our case, analyzing infinite series, we're particularly interested in the summation from \( n=1 \) to infinity.

Exponential Terms

Exponential terms include expressions like \( 10^n \) in the denominator. These terms grow extremely fast as \( n \) increases.

This rapid growth impacts the series significantly, often leading the terms towards zero when evaluated.
For instance, \( \frac{n^{10}}{10^n} \) rapidly becomes smaller as \( n \) increases, because \( 10^n \) grows more quickly than \( n^{10} \).

When you see these in a series, consider the Ratio Test, which examines the effect of such fast-growing terms on the convergence of a series.

Limit Evaluation

Limit evaluation is crucial in determining convergence. This process involves calculating the limit of the ratio of consecutive terms of a series as \( n \) approaches infinity.

In simple terms, you're checking how the series terms behave as the series grows. The outcome guides us on the convergence behavior. For the series \( \sum_{n=1}^{\infty} \frac{n^{10}}{10^n} \), we find the limit of the ratio of successive terms.

• We simplified the ratio, using limits to reveal that \( \lim_{n\to\infty} \left( \frac{1}{10} \right)^{10} = \frac{1}{10} \).

This confirms convergence as it is less than one.