Root Test

The Root Test is a method used in the calculus to the determine the convergence or divergence of the infinite series. It is particularly useful for the series where the terms involve exponential functions or factorials. The test provides the criterion based on the nth root of the terms in the series to the assess whether the series converges absolutely.

What is Root Test?

The Root Test also known as the nth Root Test is a convergence test for the infinite series. It is used to the determine whether a series converges or diverges based on the nth root of the absolute value of its terms. The test is particularly useful for the series with the terms that involve exponentials or factorials. By applying the Root Test, we can simplify the process of the evaluating the convergence of the complex series.

Statement of Root Test

To prove the Root Test, consider a series ∑a_n. The Root Test examines the limit of nth root of the absolute value of the terms:

L = \lim_{n \to \infty} \sqrt[n]{|a_n|}

Then,

• If L < 1, then ∑a_n converges absolutely.
• If L > 1 or L = ∞ then ∑a_n diverges.
• If L = 1 the test is inconclusive, and the series may converge or diverge.

Proof of Root Test

Let's discuss each case in detail.

Convergence Case (L < 1)

• Assume L < 1. This means there exists a constant c such that L = \lim_{n \to \infty} \sqrt[n]{|a_n|} < c < 1.
• Since L < c < 1 there exists an integer N such that for the all n > N \sqrt[n]{|a_n|} < c.
• Raising both sides to the power of the n we get |a_n| < c^n.

Since c < 1, cⁿ converges to the 0 as n \to \infty. Therefore, is eventually smaller than a term of the convergent geometric series with the ratio c implying that \sum a_n converges absolutely.

Divergence Case (L > 1)

• Assume L > 1. This means there exists a constant d such that L = \lim_{n \to \infty} \sqrt[n]{|a_n|} > d > 1.
• Since L > d > 1 there exists an integer N such that for the all n > N \sqrt[n]{|a_n|} > d.
• Raising both sides to the power of the n we get |a_n| > d^n.

Since d > 1, dⁿ diverges to infinity as n \to \infty. Therefore, |a_n| is eventually larger than a term of the divergent geometric series implying that \sum a_n diverges.

Inconclusive Case (L = 1)

When L = 1 the test does not provide the enough information about the behavior of the \sum a_n. The series could converge or diverge and additional tests or methods are needed to the determine its nature.

How the Root Test Works?

For a given infinite series \sum a_n the Root Test involves evaluating the following limit:

L = \lim_{n \to \infty} \sqrt[n]{|a_n|}

Where \sqrt[n]{|a_n|} denotes the n-th root of the absolute value of the n-th term a_n of the series.

Conditions for Root Test

• If L<1: The series converges absolutely.
• If L>1 or L=∞: The series diverges.
• If L=1: The test is inconclusive and series may either converge or diverge. Other tests should be used.

Applications of the Root Test

The Root Test is useful in the various scenarios including:

• Series with Exponential Terms: When dealing with the series involving eⁿor similar terms.
• Factorials in Series: For series involving the factorial terms such as the \sum_{n=1}^{\infty} \frac{n}{n!}.
• Power Series: To determine the radius of the convergence of power series.

Solved Examples on Root Test

Example 1: Series: \sum \frac{n}{2^n}

Solution:

a_n = \frac{n}{2^n}

\sqrt[n]{|a_n|} = \sqrt[n]{\frac{n}{2^n}} = \frac{\sqrt[n]{n}}{2}

L = \lim_{n \to \infty} \frac{\sqrt[n]{n}}{2} = \frac{1}{2}

Since L < 1 the series converges.

Example 2: Series: \sum \frac{2^n}{n^2}

Solution:

a_n = \frac{2^n}{n^2}

\sqrt[n]{|a_n|} = \sqrt[n]{\frac{2^n}{n^2}} = \frac{2}{\sqrt[n]{n^2}} = \frac{2}{n^{2/n}}

L = \lim_{n \to \infty} \frac{2}{n^{2/n}} = 2

Since L > 1 the series diverges.

Example 3: Series: \sum \frac{n^2}{3^n}

Solution:

a_n = \frac{n^2}{3^n}

\sqrt[n]{|a_n|} = \sqrt[n]{\frac{n^2}{3^n}} = \frac{\sqrt[n]{n^2}}{3} = \frac{n^{2/n}}{3}

L = \lim_{n \to \infty} \frac{n^{2/n}}{3} = \frac{1}{3}

Since L < 1 the series converges.

Example 4: Series: \sum \frac{5^n}{n!}

Solution:

a_n = \frac{5^n}{n!}

\sqrt[n]{|a_n|} = \sqrt[n]{\frac{5^n}{n!}} = \frac{5}{\sqrt[n]{n!}}

Using Stirling's approximation: n! \approx \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n,

\sqrt[n]{n!} \approx \frac{n}{e}

L = \lim_{n \to \infty} \frac{5}{\frac{n}{e}} = 0

Since L < 1 the series converges.

Example 5: Series: \sum \frac{n!}{4^n}

Solution:

a_n = \frac{n!}{4^n}

\sqrt[n]{|a_n|} = \sqrt[n]{\frac{n!}{4^n}} = \frac{\sqrt[n]{n!}}{4}

Using Stirling's approximation \sqrt[n]{n!} \approx \frac{n}{e}

L = \lim_{n \to \infty} \frac{\frac{n}{e}}{4} = \infty

Since L > 1 the series diverges.

Practice Questions on Root Test

Q1. Determine whether the series \sum \frac{3^n}{n^3} converges or diverges using the Root Test.

Q2. Analyze the series \sum \frac{(n+1)!}{2^n} to check its convergence or divergence.

Q3. Apply the Root Test to the series \sum \frac{2^n}{n^2 n!} and determine the behavior.

Q4. Use the Root Test to evaluate the series \sum \frac{n^n}{10^n}.

Q5. Check the convergence of the series \sum \frac{n^2}{5^n} using the Root Test.

Q6. Determine the behavior of the series \sum \frac{4^n}{n^4} using the Root Test.

Q7. Analyze whether the series \sum \frac{5^n}{n^2} converges or diverges.

Q8. Apply the Root Test to the series \sum \frac{n^3}{3^n} and state the result.

Q9. Evaluate the series \sum \frac{n!}{n^2} to determine if it converges or diverges using the Root Test.

Q10. Check the convergence of \sum \frac{6^n}{n!} using the Root Test.

Conclusion

The Root Test is a powerful tool for the determining the convergence or divergence of the infinite series particularly useful for the series involving the exponential functions and factorials. By evaluating the n^th root of the terms it provides the clear criterion for the series behavior. When the limit L is greater than 1 the series diverges when L is less than 1 the series converges. However, if L = 1 the test is inconclusive and other methods should be employed. Overall, mastering the Root Test enhances one's ability to the analyze series efficiently and effectively.