Problem 45
Question
Prove that if \(\Sigma a_{n}\) diverges and \(\Sigma b_{n}\) converges, then \(\Sigma\left(a_{n}+b_{n}\right)\) diverges.
Step-by-Step Solution
Verified Answer
If \(\Sigma a_{n}\) diverges and \(\Sigma b_{n}\) converges, then \(\Sigma(a_{n} + b_{n})\) diverges.
1Step 1: 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.
2Step 2: 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 \(L\).
3Step 3: 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 \(s_{N} = \left(\sum_{n=1}^{N} a_{n}\right) + \left(\sum_{n=1}^{N} b_{n}\right)\).
4Step 4: 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.
5Step 5: 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.
Key Concepts
Convergence of SeriesPartial SumsSeries Addition
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:
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 — some go to infinity or don’t 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’s why partial sums are important:
\[ S_N = a_1 + a_2 + \, ... \, + a_N \]
Here’s 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’t, the series diverges.
- Understanding partial sums is crucial because they tell us about the stability of the series — whether it reaches a particular balance or not.
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:
\( \Sigma(a_n + b_n) = a_1 + b_1, a_2 + b_2, \, ... \, \)
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 — 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.
Other exercises in this chapter
Problem 45
Give conditions on \(p\) that determine the convergence or divergence of \(\sum_{n=1}^{\infty} \frac{1}{n^{p}}\left(1+\frac{1}{2^{p}}+\frac{1}{3^{p}}+\cdots+\fr
View solution Problem 45
Note that $$ \begin{aligned} 1-\frac{1}{2}+& \frac{1}{3}-\frac{1}{4}+\cdots-\frac{1}{2 n} \\ &=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2 n}-\left(1+\frac{1}{2
View solution Problem 46
Use a CAS to find the first four nonzero terms in the Maclaurin series for each of the following. Check Problems 43-48 to see that you get the same answers usin
View solution Problem 46
Find the third-order Maclaurin polynomial for \((1+x)^{3 / 2}\) and bound the error \(R_{3}(x)\) if \(-0.1 \leq x \leq 0\).
View solution