Problem 72 - Essential Calculus

Question

Prove that if \(P(n)\) and \(Q(n)\) are polynomials of degree \(j\) and \(k\) respectively, then the series \(\sum_{n=1}^{\infty} \frac{P(n)}{Q(n)}\) converges if \(j

Step-by-Step Solution

Answer

The given series \(\sum_{n=1}^{\infty} \frac{P(n)}{Q(n)}\) converges if the Polynomial in the numerator \(P(n)\) is of degree less than the Polynomial in the denominator \(Q(n)\) by more than one, \(j

Step 1: Comparison Test

The comparison test will be used in this proof, which states that let \(a_n\) and \(b_n\) be sequences of nonnegative real numbers. If \(a_n\) and \(b_n\) are series with \(0 \leq a_n \leq b_n\) for all \(n\), then the convergence or divergence of \(b_n\) determines \(a_n\). If \(b_n\) converges, \(a_n\) converges, and if \(b_n\) diverges, \(a_n\) diverges.

Step 2: Using the Comparison Test for convergence

If the degree of the polynomial in the numerator \(P(n)\) is less than the degree of the polynomial in the denominator \(Q(n)\) by more than one, \(j

Step 3: Using the Comparison Test for divergence

If the degree of the polynomial in the numerator \(P(n)\) is equal to or greater than the degree of the polynomial in the denominator \(Q(n)\) minus one, \(j \geq k-1\), the series diverges. We can let and compare our series with \(1/n^{k-j}\), where \(P(n)/Q(n) \geq 1/n^{k-j}\). Because the degree of \(Q(n)\) is less than or equal to one more than the degree of \(P(n)\), our comparison series \(1/n^{k-j}\) is harmonic or worse, and hence diverges. Therefore, from the comparison test, \(P(n)/Q(n)\) diverges.

Key Concepts

Comparison Test

The Comparison Test is a vital tool in determining the convergence or divergence of series by comparing it with another series whose behavior is known. This test applies to nonnegative series, and it works by bounding one series with another. If you have two sequences, \(a_n\) and \(b_n\), where \(0 \leq a_n \leq b_n\), you can use this relationship to determine convergence or divergence.

• If the series \(\sum b_n\) converges, then \(\sum a_n\) also converges.
• Conversely, if the series \(\sum b_n\) diverges, the series \(\sum a_n\) diverges as well.

This method is particularly useful for series of rational functions, like those formed by dividing polynomials. By choosing a suitable \(b_n\), one can often simplify the convergence test for complex expressions. This is key in the problem at hand, where the focus is on comparing series with polynomials in numerator and denominator.

Polynomial Degree

Understanding polynomial degree is crucial when examining the convergence of series involving \(\frac{P(n)}{Q(n)}\) where \(P(n)\) and \(Q(n)\) are polynomials of degree \(j\) and \(k\), respectively. The degree of a polynomial is simply the highest power of the variable in the equation.

• For instance, if \(P(n) = 3n^2 + 2n + 1\), the degree is 2 because of the \(n^2\) term.
• If \(Q(n) = n^3 + 4n^2 + n + 2\), then the degree is 3.

When dealing with series like \(\sum \frac{P(n)}{Q(n)}\), the degrees of \(P(n)\) and \(Q(n)\) significantly influence convergence. If \(j < k-1\), the polynomial in the numerator grows slower than that in the denominator, leading to convergence due to increasing terms being divided by significantly larger ones. However, if \(j \geq k-1\), the growth rates balance out or favor the numerator, often leading to divergence.

Divergence of Series

In mathematics, a series diverges if the sum of its terms increases indefinitely, reaching neither a finite limit nor any form of stabilization. For the series \(\sum \frac{P(n)}{Q(n)}\), divergence occurs when the polynomial in the numerator isn't sufficiently outweighed by the polynomial in the denominator.

• When \(j \geq k-1\), meaning the degree of \(P(n)\) is at least one less than that of \(Q(n)\), the series diverges.
• This is because, asymptotically, \(P(n)\) grows to counterbalance or exceed \(Q(n)\), causing the terms of the series not to approach zero quickly enough.

An example is trying to compare with the harmonic series \(\frac{1}{n}\), which is known to diverge. Here, the divergence is flagged by the inability of the denominator to dominate the numerator's growth. Recognizing these signs is essential in applying divergence tests effectively.

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Problem 72

Other exercises in this chapter

Problem 72

State the \(n\) th-Term Test for Divergence.

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Problem 72

Determine whether the sequence with th given \(n\) th term is monotonic. Discuss the boundedness of th sequence. Use a graphing utility to confirm your results.

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Problem 72

Evaluate the binomial coefficient using the formula \(\left(\begin{array}{l}k \\\ n\end{array}\right)=\frac{k(k-1)(k-2)(k-3) \cdot \cdots(k-n+1)}{n !}\) where \

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Problem 72

Use the Root Test to determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty}\left(\frac{\ln n}{n}\right)^{n} $$

View solution

Key Concepts

Comparison Test

The Comparison Test is a vital tool in determining the convergence or divergence of series by comparing it with another series whose behavior is known. This test applies to nonnegative series, and it works by bounding one series with another. If you have two sequences, \(a_n\) and \(b_n\), where \(0 \leq a_n \leq b_n\), you can use this relationship to determine convergence or divergence.

• If the series \(\sum b_n\) converges, then \(\sum a_n\) also converges.
• Conversely, if the series \(\sum b_n\) diverges, the series \(\sum a_n\) diverges as well.

This method is particularly useful for series of rational functions, like those formed by dividing polynomials. By choosing a suitable \(b_n\), one can often simplify the convergence test for complex expressions. This is key in the problem at hand, where the focus is on comparing series with polynomials in numerator and denominator.

Polynomial Degree

Understanding polynomial degree is crucial when examining the convergence of series involving \(\frac{P(n)}{Q(n)}\) where \(P(n)\) and \(Q(n)\) are polynomials of degree \(j\) and \(k\), respectively. The degree of a polynomial is simply the highest power of the variable in the equation.

• For instance, if \(P(n) = 3n^2 + 2n + 1\), the degree is 2 because of the \(n^2\) term.
• If \(Q(n) = n^3 + 4n^2 + n + 2\), then the degree is 3.

When dealing with series like \(\sum \frac{P(n)}{Q(n)}\), the degrees of \(P(n)\) and \(Q(n)\) significantly influence convergence. If \(j < k-1\), the polynomial in the numerator grows slower than that in the denominator, leading to convergence due to increasing terms being divided by significantly larger ones. However, if \(j \geq k-1\), the growth rates balance out or favor the numerator, often leading to divergence.

Divergence of Series

In mathematics, a series diverges if the sum of its terms increases indefinitely, reaching neither a finite limit nor any form of stabilization. For the series \(\sum \frac{P(n)}{Q(n)}\), divergence occurs when the polynomial in the numerator isn't sufficiently outweighed by the polynomial in the denominator.

• When \(j \geq k-1\), meaning the degree of \(P(n)\) is at least one less than that of \(Q(n)\), the series diverges.
• This is because, asymptotically, \(P(n)\) grows to counterbalance or exceed \(Q(n)\), causing the terms of the series not to approach zero quickly enough.

An example is trying to compare with the harmonic series \(\frac{1}{n}\), which is known to diverge. Here, the divergence is flagged by the inability of the denominator to dominate the numerator's growth. Recognizing these signs is essential in applying divergence tests effectively.