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Chapter 9: Problem 45

Prove that if \(\Sigma a_{n}\) diverges and \(\Sigma b_{n}\) converges, then \(\Sigma\left(a_{n}+b_{n}\right)\) diverges.

Short Answer

Expert verified
If \(\Sigma a_{n}\) diverges and \(\Sigma b_{n}\) converges, then \(\Sigma(a_{n} + b_{n})\) diverges.

Step by step solution

01

Understand the Problem

The problem asks us to prove that if a series \(\Sigma a_{n}\) diverges and another series \(\Sigma b_{n}\) converges, then the series \(\Sigma(a_{n} + b_{n})\) also diverges. In other words, adding a convergent series to a divergent series will result in a divergent series.
02

Define Divergence and Convergence

Recall that a series \(\Sigma a_{n}\) diverges if it does not approach a finite limit, while \(\Sigma b_{n}\) converges if its partial sums approach a finite limit, say \(L\).
03

Analyze the Combined Series

Consider the partial sums of \(\Sigma(a_{n} + b_{n})\), which are \(s_{N} = \sum_{n=1}^{N} (a_{n}+b_{n})\). These can be expressed as \(s_{N} = \left(\sum_{n=1}^{N} a_{n}\right) + \left(\sum_{n=1}^{N} b_{n}\right)\).
04

Use Properties of the Series

Since \(\Sigma b_{n}\) converges, as \(N\) approaches infinity, \(\sum_{n=1}^{N} b_{n}\) approaches \(L\). However, since \(\Sigma a_{n}\) diverges, \(\sum_{n=1}^{N} a_{n}\) cannot approach any limit.
05

Prove the Divergence of the Combined Series

For \(s_{N} = \left(\sum_{n=1}^{N} a_{n}\right) + L\) as \(N\) approaches infinity, the term \(\sum_{n=1}^{N} a_{n}\) does not approach a limit. Therefore, \(s_{N}\) also does not approach a limit, leading to the conclusion that \(\Sigma(a_{n} + b_{n})\) diverges.

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Key Concepts

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

Convergence of Series
Convergence deals with the idea of a sequence or series settling down to a specific value. When we talk about the convergence of a series, we focus on its partial sums reaching a certain finite number as the number of terms increases. Suppose we have a series represented by the sum: \( \Sigma b_n \). This series is said to converge if, as we add more and more terms of the sequence \( b_n \), the total or the partial sum will reach a limit, which we'll call \( L \). It doesn't matter how many terms we add, just that after adding enough, the successive sums come closer and closer to \( L \).

Here are a few key points to remember about convergence:
  • The limit \( L \) is fixed and finite, meaning it won't change as we add more terms.
  • A convergent series "stabilizes" to a sum, so adding more terms beyond a certain point barely changes the total sum.
  • Not all series converge 鈥� some go to infinity or don鈥檛 settle to any specific value.
Partial Sums
Partial sums are the building blocks of understanding series. Let's start by saying you have a sequence, and now you want to add up its terms to potentially get a series. By adding a finite number of terms from the start of the sequence, you arrive at what's known as a partial sum. For a sequence \( \{a_n\} \), if you add the first \( N \) terms, you get the \( N \)th partial sum. This is mathematically represented as:
\[ S_N = a_1 + a_2 + \, ... \, + a_N \]
Here鈥檚 why partial sums are important:
  • They help us explore whether a series converges or diverges. By examining the behavior as \( N \) grows, we learn a lot about the series itself.
  • If the series converges, the partial sums settle to a finite number. Conversely, if they don鈥檛, the series diverges.
  • Understanding partial sums is crucial because they tell us about the stability of the series 鈥� whether it reaches a particular balance or not.
In our main example, we look at two series: \( \Sigma a_n \) and \( \Sigma b_n \), each with their own partial sums that we need to analyze for convergence or divergence.
Series Addition
Series addition involves combining two or more series to form a new series. This is what we do when we look at the series \( \Sigma(a_n + b_n) \). We are effectively adding each term of the first series \( a_n \) to each term of the second series \( b_n \), and then considering their sum:
\( \Sigma(a_n + b_n) = a_1 + b_1, a_2 + b_2, \, ... \, \)
This exercise investigates how series addition impacts the behavior of convergence or divergence:
  • Even if one of the original series such as \( \Sigma a_n \) diverges, it significantly affects the overall behavior of the combined series.
  • If \( \Sigma b_n \) converges to a limit \( L \), each of its partial sums approaches \( L \).
  • However, if adding these terms still results in divergent behavior 鈥� no single limit \( L \) for the combined partial sums \( s_N = \Sigma(a_n + b_n) \), it indicates the entire new series \( \Sigma(a_n + b_n) \) diverges.
  • This highlights how the divergence of one part in a combination is enough to lead the whole new series into divergence.
To sum up, understanding how series addition affects convergence is essential in tackling various mathematical problems involving series.

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Most popular questions from this chapter

In Problems 21-30, find an explicit formula \(a_{n}=\) for each sequence, determine whether the sequence converges or diverges, and, if it converges, find \(\lim _{n \rightarrow \infty} a_{n}\). \(1, \frac{1}{1-\frac{1}{2}}, \frac{1}{1-\frac{2}{3}}, \frac{1}{1-\frac{3}{4}}, \ldots\)

In Problems 9-28, find the convergence set for the given power series. Hint: First find a formula for the nth term; then use the Absolute Ratio Test. $$ x+2^{2} x^{2}+3^{2} x^{3}+4^{2} x^{4}+\cdots $$

In Problems 21-30, find an explicit formula \(a_{n}=\) for each sequence, determine whether the sequence converges or diverges, and, if it converges, find \(\lim _{n \rightarrow \infty} a_{n}\). \(1, \frac{2}{2^{2}-1^{2}}, \frac{3}{3^{2}-2^{2}}, \frac{4}{4^{2}-3^{2}}, \ldots\)

In Problems \(1-8\), find the convergence set for the given power series. $$ \sum_{n=1}^{\infty}(-1)^{n} \frac{(x-2)^{n}}{n} $$

In Problems 21-30, find an explicit formula \(a_{n}=\) for each sequence, determine whether the sequence converges or diverges, and, if it converges, find \(\lim _{n \rightarrow \infty} a_{n}\). \(\frac{1}{2-\frac{1}{2}}, \frac{2}{3-\frac{1}{3}}, \frac{3}{4-\frac{1}{4}}, \frac{4}{5-\frac{1}{5}}, \ldots\)
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