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Chapter 10: Problem 68

Does the series $$\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n^{2}}\right)$$ converge or diverge? Justify your answer.

Short Answer

Expert verified
The series diverges.

Step by step solution

01

Identify the Series Terms

The series given is \( \sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{n^2} \right) \). We can express the general term of the series as \( a_n = \frac{1}{n} - \frac{1}{n^2} \).
02

Simplify the General Term

Simplifying \( a_n = \frac{1}{n} - \frac{1}{n^2} \), we note that this can be rewritten as a single fraction: \( a_n = \frac{n-1}{n^2} \).
03

Test for Comparison with a Known Series

Observe that \( a_n = \frac{n-1}{n^2} \leq \frac{1}{n} \) for all \( n \geq 1 \). Since \( \sum \frac{1}{n} \) is the harmonic series, it is known to diverge.
04

Apply the Limit Comparison Test

To further justify, we apply the limit comparison test comparing \( a_n \) with \( b_n = \frac{1}{n} \):\[ \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\frac{n-1}{n^2}}{\frac{1}{n}} = \lim_{n \to \infty} \frac{n-1}{n} = \lim_{n \to \infty} \left(1 - \frac{1}{n} \right) = 1. \]Since the limit is a positive finite number, both series \( \sum_{n=1}^{\infty} a_n \) and \( \sum_{n=1}^{\infty} b_n \) either converge or diverge together.
05

Conclusion on Convergence or Divergence

Since \( \sum_{n=1}^{\infty} \frac{1}{n} \) diverges and by the limit comparison test, \( \sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{n^2} \right) \) also diverges.

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Key Concepts

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

Limit Comparison Test
The limit comparison test is a handy method used in calculus to decide whether a series converges or diverges. It's particularly useful when you want to compare two series. If we have two series, \( \sum a_n \) and \( \sum b_n \), and we believe they exhibit similar behavior, we can use this test to confirm our hypothesis.
  • Pick two series: \( \sum a_n \) is the series you want to analyze, and \( \sum b_n \) is a series you know about.
  • Compute the limit: \( \lim_{{n \to \infty}} \frac{a_n}{b_n} \).
  • If this limit is a positive finite number, then both series either converge or diverge together.
In our exercise, we selected the well-known harmonic series \( \sum \frac{1}{n} \) as the comparison, \( b_n \). Calculating:\[\lim_{{n \to \infty}} \frac{\frac{n-1}{n^2}}{\frac{1}{n}} = \lim_{{n \to \infty}} \left( 1 - \frac{1}{n} \right) = 1.\]Since this limit equals 1, a positive number, the denominated series and the harmonic series will both diverge.
Harmonic Series
The harmonic series is one of the best-known divergent series in mathematics. It is written as \( \sum_{n=1}^{\infty} \frac{1}{n} \).
  • Each term of the series is the reciprocal of an integer.
  • While the terms get smaller, \( \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \)..., infinitely, they never reach zero.
  • This characteristic leads the harmonic series to diverge.
Despite the small size of the terms as \( n \) increases, the sum grows without bound. The diverging nature of this series is often used for comparisons, typically as a benchmark for other series when applying various tests for convergence or divergence.
Divergent Series
A divergent series is one that doesn't settle on a finite limit. This means as you keep adding more terms, the total sum keeps increasing beyond any boundary. It doesn鈥檛 approach a fixed number.
  • This is the opposite of a convergent series, which settles on a specific value.
  • Series like the harmonic series are classic examples.
  • To check if a series diverges, applying tests such as the ratio, root, or comparison tests can be effective strategies.
In our original exercise, the limit comparison test proved that the given series behaves like the harmonic series, which is known to diverge. Hence, itself is also a divergent series, reinforcing its perpetual growth without limit.

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Most popular questions from this chapter

Which of the series converge, and which diverge? Use any method, and give reasons for your answers. \begin{equation}\sum_{n=1}^{\infty} \frac{2^{n}-n}{n 2^{n}}\end{equation}

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Use the identity \(\sin ^{2} x=(1-\cos 2 x) / 2\) to obtain the Maclaurin series for \(\sin ^{2} x .\) Then differentiate this series to obtain the Maclaurin series for 2 \(\sin x \cos x .\) Check that this is the series for \(\sin 2 x .\)
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