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Chapter 8: Problem 57

Use a Comparison Test to determine whether the given series converges or diverges. $$ \sum_{n=2}^{\infty} \frac{n}{\ln (n)^{3}} $$

Short Answer

Expert verified
The series \( \sum_{n=2}^{\infty} \frac{n}{\ln(n)^3} \) diverges.

Step by step solution

01

Identify the Series to Compare

To use the Comparison Test, we need another series whose behavior we know. The given series is \( \sum_{n=2}^{\infty} \frac{n}{\ln(n)^3} \). We can compare it to the series \( \sum_{n=2}^{\infty} \frac{1}{n} \), which is known to diverge. This is because \( \ln(n)^3 > 1 \) for all \( n \geq 2 \).
02

Set Up the Comparison Inequality

We need to compare \( \frac{n}{\ln(n)^3} \) with \( \frac{1}{n} \). For comparison, notice: \[ \frac{n}{\ln(n)^3} > \frac{1}{n} \iff n^2 > \ln(n)^3. \] This inequality must be checked for values of \( n \geq 2 \), which generally holds true for large \( n \).
03

Apply the Direct Comparison Test

Since \( \frac{n}{\ln(n)^3} > \frac{1}{n} \) for large \( n \) and \( \sum_{n=2}^{\infty} \frac{1}{n} \) diverges, by the Direct Comparison Test, the series \( \sum_{n=2}^{\infty} \frac{n}{\ln(n)^3} \) also diverges.

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Key Concepts

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

Understanding Series Convergence
The notion of a series converging or diverging is fundamental in calculus and mathematical analysis. Series convergence means that as you add more and more terms, the total sum approaches a fixed value鈥攃alled the limit. This concept is crucial for analyzing the behavior of sequences and ensuring stable results in computations.
When a series diverges, it means the sum does not settle down to a particular number. Instead, it either increases or decreases without bound as more terms are added.
  • A convergent series: The sum gets closer to a specific value.
  • A divergent series: The sum increases or decreases indefinitely.
Identifying whether a series converges or diverges helps in many fields where predicting behavior over long terms is necessary, like physics and finance.
Explaining the Direct Comparison Test
The Direct Comparison Test is a powerful tool to determine the convergence or divergence of a series. The main idea is straightforward: compare the given series with another series whose convergence or divergence is already known.
To use this test, follow these steps:
  • Identify another series that is simpler and well-known, such as the harmonic series or a geometric series.
  • Show that each term of the given series is either smaller or greater than the corresponding term of the known series.
  • Apply the rules of the test:
    • If the known series converges and the terms of your series are smaller, then your series converges.
    • If the known series diverges and the terms of your series are greater, then your series diverges.
This method provides a clear, step-by-step way to analyze series and make conclusions about their behavior.
The Divergence of the Harmonic Series
The harmonic series is a well-known example of a divergent series and is crucial in understanding why some series don't converge to a limit. It is represented as \[ \sum_{n=1}^{\infty} \frac{1}{n} \]The key characteristic of the harmonic series is that its terms get progressively smaller, but not quite fast enough for the sum to level off to a fixed value.
Here is why the harmonic series diverges:
  • The sum of the terms approaches infinity as more terms are added despite each term getting smaller.
  • Its partial sums keep increasing beyond any particular bound.
Understanding the divergence of the harmonic series allows us to use it as a baseline or comparison when evaluating other series, as we see with the Direct Comparison Test.

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Most popular questions from this chapter

Suppose that every dollar that we spend gives rise (through wages, profits, etc.) to 90 cents for someone else to spend. That 90 cents will generate a further 81 cents for spending, and so on. How much spending will result from the purchase of a \(\$ 16,000\) automobile, the car included? (This phenomenon is known as the multiplier effect.)

Use a Taylor polynomial with base point \(c=e^{3}\) to approximate \(\ln (20)\) to five decimal places. Do not use a calculator to evaluate any value of \(\ln (x),\) but you may use a calculator for arithmetic with the number \(e\) and its powers.

Use partial fractions to calculate the \(N^{\text {th }}\) partial sum \(S_{N}\) of the given series in closed form. Sum the series by finding \(\lim _{N \rightarrow \infty} S_{N}\). $$ \sum_{n=1}^{\infty} \frac{2 n+1}{\left(n^{2}+n\right)^{2}} $$

If \(f(x)=\sum_{n=0}^{\infty} a_{n} x^{n}\) and \(g(x)=\sum_{n=0}^{\infty} b_{n} x^{n}\) converge on \((-R, R),\) then we may formally multiply the series as though they were polynomials. That is, if \(h(x)=f(x) g(x),\) then $$ h(x)=\sum_{n=0}^{\infty}\left(\sum_{k=0}^{n} a_{k} b_{n-k}\right) x^{n} $$ The product series, which is called the Cauchy product, also converges on \((-R, R) .\) Exercises \(59-62\) concern the Cauchy product. Suppose that the series \(\sum_{n=0}^{\infty} a_{n} x^{n}\) converges on \((-R, R)\) to a function \(f(x)\) and that \(|f(x)| \geq k>0\) on that interval for some positive constant \(k\). Then, \(1 / f(x)\) also has a convergent power series expansion on \((-R, R) .\) Compute its coefficients in terms of the \(a_{n}\) 's. Hint: Set $$ \frac{1}{f(x)}=g(x)=\sum_{n=0}^{\infty} b_{n} x^{n} $$ Use the equation \(f(x) \cdot g(x)=1\) to solve for the \(b_{n}\) 's.

The inequality $$ \frac{3 \sin (t)}{2+\cos (t)}0) $$ was discovered by Nicholas Cusa in \(1458 .\) Calculate the Taylor polynomial of degree 5 for \(f(t)=3 \sin (t) /(2+\cos (t))\) about \(c=0 .\) Use it to explain Cusa's inequality for small positive values of \(t\).
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