Problem 61 - Thomas Calculus

Question

Show that \(\sum_{n=2}^{\infty}\left((\ln n)^{q} / n^{p}\right)\) converges for \(-\infty < q <\infty\) and \(p >1.\) (Hint: Limit Comparison with \(\sum_{n=2}^{\infty} 1 / n^{r}\) for \(1 < r < p.)\)

Step-by-Step Solution

Answer

The series converges for all real \( q \) and \( p > 1 \).

Step 1: Identify Similar Series

We are given the series \( \sum_{n=2}^{\infty} \frac{(\ln n)^{q}}{n^{p}} \). The hint directs us to compare this series with \( \sum_{n=2}^{\infty} \frac{1}{n^{r}} \) where \( 1 < r < p \). This suggests using the limit comparison test.

Step 2: Set Up the Limit Comparison

For Limit Comparison Test, calculate \( \lim_{{n \to \infty}} \frac{\frac{(\ln n)^{q}}{n^{p}}}{\frac{1}{n^{r}}} = \lim_{{n \to \infty}} \frac{(\ln n)^{q} n^{r-p}}{1} = \lim_{{n \to \infty}} (\ln n)^{q} n^{r-p} \).

Step 3: Evaluate the Limit

We analyze \( \lim_{{n \to \infty}} (\ln n)^{q} n^{r-p} \). As \( n \to \infty \), if \( r < p \), then \( n^{r-p} \to 0 \). Thus, \( \lim_{{n \to \infty}} (\ln n)^{q} \cdot 0 = 0 \).

Step 4: Check Conditions for Convergence

The limit found is zero, which implies convergence of the original series by the limit comparison test since \( \sum_{n=2}^{\infty} \frac{1}{n^{r}} \) is convergent for \( r > 1 \). Thus, \( \sum_{n=2}^{\infty} \frac{(\ln n)^{q}}{n^{p}} \) converges.

Key Concepts

Convergence of series

Understanding the convergence of a series is fundamental in mathematical analysis. A series is said to converge when the sum of its infinite terms approaches a specific value. This tells us that as you continue adding terms of the series, you approach a limit rather than diverging off to infinity. It's important to identify when a series converges because it allows mathematicians to handle infinite processes more comfortably.

In the given exercise, we examine the series:

• \( \sum_{n=2}^{\infty} \frac{(\ln n)^{q}}{n^{p}} \).

To determine convergence, we use tests like the Limit Comparison Test, which compares our series to a known convergently-behaving series, such as:

• \( \sum_{n=2}^{\infty} \frac{1}{n^{r}} \)

for certain conditions on \( r \), helping us explore the series' behavior at infinity. If our given series and a comparative series both behave similarly at infinity and the comparative series is known to converge, we can conclude that the original one converges as well.

Logarithmic function

Logarithmic functions, like \( \ln n \), play an essential role in various mathematical series, particularly when exploring their limits and convergence. The natural logarithm, \( \ln \), relates to the exponential function \( e \) in that it answers the question: what power must \( e \) be raised to in order to obtain \( n \)? It's a fundamental function in mathematical analysis for dealing with exponential growth or decay contexts.

In the series:

• \( \sum_{n=2}^{\infty} \frac{(\ln n)^{q}}{n^{p}} \)

\( (\ln n)^{q} \) introduces a complexity that requires careful consideration when evaluating the convergence. Since natural logarithms grow at a slower pace than polynomial functions, they affect the series' convergence depending on their exponent, \( q \). Whether \( q \) is positive, negative or zero, it can change how dominant the term \( (\ln n)^{q} \) is compared to \( n^{p} \) at infinity. However, the given conditions in the exercise (with appropriate \( p \) and \( q \)) ensure that this component does not inhibit convergence.

Mathematical series analysis

Analyzing a mathematical series involves many techniques to discover important properties like convergence, divergence, and bounds. Series analysis helps in the sum evaluation of functions over an infinite domain—an essential component for countless applications in science, engineering, and mathematics.

The given series:

• \( \sum_{n=2}^{\infty} \frac{(\ln n)^{q}}{n^{p}} \)

is analyzed using the Limit Comparison Test. This test is useful for series where direct convergence tests, like the Integral Test or Ratio Test, might be cumbersome. It focuses on the ratio of the terms between two series, simplifying the analysis.

The Limit Comparison Test can establish convergence by examining the limit:

• \( \lim_{n \to \infty} \frac{\frac{(\ln n)^{q}}{n^{p}}}{\frac{1}{n^{r}}} = \lim_{n \to \infty} (\ln n)^{q} n^{r-p} \)

This ultimately leads to understanding how \( n^{r-p} \), with \( r

Previous

Problem 61

Next

Problem 61

Other exercises in this chapter

Problem 60

Which series in Exercises \(49-68\) converge, and which diverge? Give reasons for your answers. If a series converges, find its sum. $$ \sum_{n=1}^{\infty}\left

View solution

Problem 61

Obtain the Taylor series for 1\(/(1+x)^{2}\) from the series for \(-1 /(1+x) .\)

View solution

Problem 61

Convergence or Divergence Which of the series in Exercises \(55-62\) converge, and which diverge? Give reasons for your answers. $$\sum_{i=1}^{\infty} \frac{1 \

View solution

Problem 62

Use the Taylor series for 1\(/\left(1-x^{2}\right)\) to obtain a series for 2\(x /\left(1-x^{2}\right)^{2}\)

View solution