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Use the Ratio Test to determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty} \frac{2^{n}}{n^{5}} $$

Short Answer

Expert verified
The series \( \sum_{n=1}^{\infty} \frac{2^{n}}{n^{5}} \) diverges.

Step by step solution

01

Identify the series

For the series \( \sum_{n=1}^{\infty} \frac{2^{n}}{n^{5}} \) the \( a_n = \frac{2^{n}}{n^{5}} \).
02

Calculate the limit L

To apply the Ratio Test, calculate the limit \( L = \lim_{n \rightarrow \infty} | \frac{a_{n+1}}{a_n} | \). Substitute \( a_{n+1} = \frac{2^{n+1}}{(n+1)^{5}} \) and \( a_n = \frac{2^{n}}{n^{5}} \) into the formula: \[ \lim_{n \rightarrow \infty} \frac{2^{n+1}}{(n+1)^{5}} * \frac{n^{5}}{2^{n}} = \lim_{n \rightarrow \infty} \frac{2n^{5}}{(n+1)^{5}} \] which simplifies to \[ \lim_{n \rightarrow \infty} \frac{2}{((n+1)/n)^{5}} \]
03

Evaluate the limit

As n approaches infinity, the expression ((n+1)/n) approaches 1, therefore, the denominator becomes 1 and \( L = 2 \)
04

Applying Ratio Test

Because L is greater than 1, by the Ratio Test, the given series diverges.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Infinite Series Convergence
When we examine an infinite series, a sum in which the number of terms increases without bound, our fundamental question is whether it converges to a finite value or diverges, growing indefinitely. This distinction is crucial when using infinite series to model real-world phenomena or when trying to understand the long-term behavior of sequences in mathematics.

The Ratio Test is particularly useful to determine the convergence of an infinite series that consists of terms involving factorials, exponential functions, and other types that might prove challenging with other tests. The test compares the limit of the absolute value of consecutive terms of the series; if this limit (usually called 'L') exists.

When applying the Ratio Test, three possible outcomes help us to draw conclusions about an infinite series:
  • If L < 1, the series converges absolutely.
  • If L > 1 or L is infinite, the series diverges.
  • If L = 1, the test is inconclusive, and the series may converge or diverge.

For example, in the given exercise, the series \( \sum_{n=1}^{\infty} \frac{2^{n}}{n^{5}} \) was shown to diverge by the Ratio Test because the limit 'L' of the ratio of successive terms was found to be greater than 1.
Limit of a Sequence
The limit of a sequence is a fundamental concept in calculus, which defines the value that the terms of a numerical sequence 'approach' as the index (usually denoted as 'n') goes to infinity. If the terms of the sequence become arbitrarily close to a single number L as 'n' becomes very large, we say that L is the limit of the sequence.

Mathematically, we express this using the notation \(\lim_{n \to \infty} a_n = L\), where \(a_n\) is the nth term of the sequence. The concept of a limit is crucial for understanding the behavior of sequences and, by extension, series when looking at an infinite number of terms.

When calculating limits for the Ratio Test as seen in step 2 and 3 of our exercise, simplifying the expression before taking the limit can often make the process easier. For the given series, the limit was simplified to \(\lim_{n \to \infty} \frac{2}{((n+1)/n)^{5}}\), highlighting the exponential nature of the numerator relative to the polynomial nature of the denominator.
Exponential Functions
An exponential function is a mathematical expression in which a constant base is raised to a variable exponent. Typically written in the form \( b^{x} \) where 'b' is the base and 'x' is the exponent, exponential functions grow (or decay) at rates proportional to their current value.

These functions are represented by rapidly increasing (or decreasing) curves on a graph and are essential in modeling growth processes, such as population growth, radioactive decay, and interest compounding in finance.

In the context of the Ratio Test in our exercise, the exponential function is represented by \( 2^{n} \) and this exponential growth in the numerator of the series' terms plays a pivotal role in determining divergence. Unlike polynomial growth, as seen in the denominator with \( n^{5} \) where the degree of growth is fixed, exponential growth accelerates without bound, leading to a limit 'L' greater than 1 and therefore a divergent series according to the Ratio Test.

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