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Prove that an absolutely convergent series \(\sum_{n=1}^{\infty} a_{n}\) is convergent. Hint: Put \(b_{n}=\) \(a_{n}+\left|a_{n}\right| .\) Then the \(b_{n}\) are nonnegative; we have \(\left|b_{n}\right| \leq 2\left|a_{n}\right|\) and \(a_{n}=b_{n}-\left|a_{n}\right|.\)

Short Answer

Expert verified
An absolutely convergent series \(\sum_{n=1}^{\infty}a_{n}\) converges.

Step by step solution

01

Define Absolute Convergence

An absolutely convergent series is one where the series \(\sum_{n=1}^{\infty}\left|a_{n}\right|\) converges, meaning the sum of the absolute values of the terms is finite.
02

Define the New Sequence

Set \(b_{n} = a_{n} + |a_{n}|\). Notice that this makes each \(b_{n}\) nonnegative because \(a_{n}\) and \(|a_{n}|\) always sum to a nonnegative value.
03

Bound the New Sequence

Show that \(|b_{n}| \leq 2|a_{n}|\). Since \(|a_{n}|\) is the positive value of \(a_{n}\), the absolute value of \(b_{n}\) is at most twice the absolute value of \(a_{n}\).
04

Express the Original Sequence

Express \(a_{n}\) in terms of the new sequence \(b_{n}\):\(a_{n} = b_{n} - |a_{n}|\). Because \(b_{n}\) is nonnegative, this expression holds.
05

Show Convergence of the New Series

Since \(\sum_{n=1}^{\infty}|a_{n}|\) converges and \(|b_{n}| \leq 2|a_{n}|\), it follows from the Comparison Test that the series \(\sum_{n=1}^{\infty}b_{n}\) converges too.
06

Conclude Absolute Convergence Implies Convergence

Finally, because \(\sum_{n=1}^{\infty}b_{n}\) converges and \(|a_{n}|\) converges, the sum of \(\sum_{n=1}^{\infty}a_{n}\) converges too, proving that absolutely convergent series are convergent.

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Key Concepts

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

Series Convergence
When we talk about series convergence, we are concerned with whether a series adds up to a finite value as we sum more and more terms. Take a series \( \sum_{n=1}^{\infty} a_{n} \). The basic idea is if you keep adding the terms together, do you approach a specific number? If you do, then the series converges. This is an essential concept because it helps predict the behavior of the series as it extends to infinity. For a series to be convergent, the terms must get smaller in such a manner that their sum approaches a finite limit. Otherwise, the series is divergent and doesn't sum to a finite value.
Comparison Test
The Comparison Test is a powerful tool for determining the convergence of a series. Here's how it works: If you have two series \( \sum_{n=1}^{\infty} a_{n} \) and \( \sum_{n=1}^{\infty} b_{n} \) with non-negative terms, and if \( 0 \leq a_{n} \leq b_{n} \) for all \( n \), then the following holds:
  • If \( \sum_{n=1}^{\infty} b_{n} \) converges, then \( \sum_{n=1}^{\infty} a_{n} \) also converges.
  • If \( \sum_{n=1}^{\infty} a_{n} \) diverges, then \( \sum_{n=1}^{\infty} b_{n} \) also diverges.
This test is useful because sometimes the direct computation of the sum of a series is difficult. By comparing it to another series whose behavior we understand better, we can infer convergence or divergence. In our exercise, we use the Comparison Test to show that the series of \( b_{n} \) converges because it's bounded by a convergent series of \( 2 \left| a_{n} \right| \).
Sequence Bound
In mathematics, defining bounds is crucial for understanding the behavior of sequences and series. A sequence is said to be bounded if there exists a number that is larger than or equal to every term in the sequence. For instance, if \( |a_n| \) represents the absolute value of the terms in a sequence, then having an upper bound means there is some constant M such that:
\[ |a_n| \leq M, \text{ for all } n. \]
In our exercise, we used the sequence bound concept by introducing a new sequence \( b_n = a_n + |a_n| \). This new sequence is non-negative and is bounded by \( 2|a_n| \), which means:
\[ |b_n| \leq 2|a_n|. \]
Because the series of \( |a_n| \) converges, it implies the series of \( b_n \) converges too, based on our previous explanation with the Comparison Test. Bounding helps simplify and solve more complex problems as it provides a manageable way to compare and analyze sequences.

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Most popular questions from this chapter

Use Maclaurin series to do and check your results by computer. $$\left.\frac{d^{6}}{d x^{6}}\left(x^{4} e^{x^{2}}\right)\right|_{x=0}$$

Find the first few terms of the Maclaurin series for each of the following functions and check your results by computer. $$\ln \left(2-e^{-x}\right)$$

Test the following series for convergence or divergence. Decide for yourself which test is easiest to use, but don't forget the preliminary test. Use the facts stated above when they apply. $$\sum_{n=1}^{\infty} \frac{n^{n}}{n !}$$

(a) It is clear that you (or your computer) can't find the sum of an infinite series just by adding up the terms one by one. For example, to get \(\zeta(1.1)=\) \(\sum_{n=1}^{\infty} 1 / n^{1.1}(\text { see Problem } 15.22)\) with error \(<0.005\) takes about \(10^{33}\) terms. To see a simple alternative (for a series of positive decreasing terms) look at Figures 6.1 and 6.2. Show that when you have summed \(N\) terms, the sum \(R_{N}\) of the rest of the series is between \(I_{N}=\int_{N}^{\infty} a_{n} d n\) and \(I_{N+1}=\int_{N+1}^{\infty} a_{n} d n\) (b) Find the integrals in (a) for the \(\zeta(1.1)\) series and verify the claimed number of terms needed for error \(<0.005 .\) Hint: Find \(N\) such that \(I_{N}=0.005 .\) Also find upper and lower bounds for \(\zeta(1.1)\) by computing \(\sum_{n=1}^{N} 1 / n^{1.1}+\int_{N}^{\infty} n^{-1.1} d n\) and \(\sum_{n=1}^{N} 1 / n^{1.1}+\int_{N+1}^{\infty} n^{-1.1} d n\) where \(N\) is far less than \(10^{33} .\) Hint: You want the difference between the upper and lower limits to be about \(0.005 ;\) find \(\mathrm{N}\) so that term \(a_{N}=0.005\).

Use the ratio test to find whether the following series converge or diverge: $$\sum_{n=0}^{\infty} \frac{(n !)^{3} e^{3 n}}{(3 n) !}$$

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