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Suppose that \(a_{n}\) is a sequence of positive numbers and the sequence \(S_{n}\) of partial sums of \(a_{n}\) is bounded above. Explain why \(\sum_{n=1}^{\infty} a_{n}\) converges. Does the conclusion remain true if we remove the hypothesis \(a_{n} \geq 0 ?\)

Short Answer

Expert verified
The series converges since \(S_n\) is bounded above and non-decreasing. Without \(a_n \geq 0\), the convergence may not hold.

Step by step solution

01

Understanding the Partial Sums

The sequence \( S_n \) of partial sums is defined as \( S_n = a_1 + a_2 + \cdots + a_n \). This sequence is given to be bounded above, meaning that there's some real number \( M \) such that \( S_n \leq M \) for all \( n \).
02

Convergence Criterion for Series

A series \( \sum_{n=1}^{\infty} a_{n} \) converges if the sequence of its partial sums \( S_n \) converges to a finite limit. Since \( S_n \) is bounded and also monotonic (since each \( a_n \geq 0 \)), the series converges by the Monotone Convergence Theorem.
03

Condition on Positive Terms

The condition \( a_n \geq 0 \) is crucial here. If this condition is removed, partial sums \( S_n \) may not maintain monotonicity (non-decreasing behavior). Thus, the bounded and convergent conclusion may fail without positivity, as negative terms could cause the partial sums to oscillate or diverge.

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Key Concepts

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

Partial Sums
In mathematics, when dealing with series, we often encounter the concept of partial sums. A partial sum is simply the sum of the first few terms of a sequence. For instance, if you have a sequence of numbers such as \(a_1, a_2, a_3, \ldots\), then the partial sum \(S_n\) is the sum of the first \(n\) terms, i.e., \(S_n = a_1 + a_2 + \ldots + a_n\).
Partial sums help us analyze the behavior of an infinite series. By examining the sequence of partial sums, we can determine whether an entire series converges or diverges. In other words, if the sequence of partial sums approaches a specific value as \(n\) increases, the series is said to be convergent.
  • Partial sums are used to simplify complex series.
  • They break an infinite process into manageable parts.
  • If a series converges, the sequence of its partial sums must settle to a single value.
Understanding partial sums is crucial to exploring various convergence tests and ensuring we know the overall behavior of a series.
Monotone Convergence Theorem
The Monotone Convergence Theorem is a powerful tool in the world of mathematics, particularly when dealing with infinite series and integrals. This theorem states that if a sequence is both monotonic and bounded, then it is guaranteed to converge.
A sequence can be monotonic in two ways: it can either be increasing (non-decreasing) or decreasing (non-increasing). For example, an increasing sequence is one where each term is greater than or equal to the previous one.
  • If a sequence \(S_n\) is increasing and bounded above, it converges to its supremum.
  • The sequence converges because it is constantly approaching a limit but never exceeding a certain boundary.
In the context of series, if we have a sequence of partial sums \(S_n\) of positive terms \(a_n\), and \(S_n\) is bounded and increasing, then by the Monotone Convergence Theorem, the series converges. It ensures that the partial sums do not just rise indefinitely but actually approach a finite value. This theorem is essential in verifying convergence when conditions like positivity of terms are met.
Bounded Sequences
A bounded sequence is a sequence of numbers that is limited in how large or small its values can get. It means there is a real number, \(M\), such that every term in the sequence is less than or equal to \(M\).
For instance, if you have a sequence \(b_n\), it is bounded above by \(M\) if for every \(n\), \(b_n \leq M\). Similarly, it can be bounded below if there's a number \(L\) such that \(b_n \geq L\) for all \(n\). A sequence that is both bounded above and below is called simply "bounded."
  • Bounded sequences provide limits to the potential size of terms.
  • The concept is crucial in determining convergence behavior.
  • Without boundedness, a sequence could potentially diverge.
In the analysis of series, if the sequence of partial sums \(S_n\) is bounded, it cannot grow indefinitely, aiding in determining whether the series converges. Boundedness, often combined with monotonicity, plays a pivotal role in applying the Monotone Convergence Theorem to ensure that we have a convergent series.

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Most popular questions from this chapter

A fast computer can sum one million terms per second of the divergent series \(\sum_{n=2}^{N} \frac{1}{n \ln n} .\) Use the integral test to approximate how many seconds it will take to add up enough terms for the partial sum to exceed 100 .

The following advanced exercises use a generalized ratio test to determine convergence of some series that arise in particular applications when tests in this chapter, including the ratio and root test, are not powerful enough to determine their convergence. The test states that if \(\lim _{n \rightarrow \infty} \frac{a_{2 n}}{a_{n}}<1 / 2\), then \(\sum a_{n}\) converges, while if \(\lim _{n \rightarrow \infty} \frac{a_{2 n+1}}{a_{n}}>1 / 2\), then \(\sum a_{n}\) diverges. Let \(a_{n}=\frac{1}{1+x} \frac{2}{2+x} \cdots \frac{n}{n+x} \frac{1}{n}=\frac{(n-1) !}{(1+x)(2+x) \cdots(n+x)} .\) Show that \(a_{2 n} / a_{n} \leq e^{-x / 2} / 2 .\) For which \(x>0\) does the generalized ratio test imply convergence of \(\sum_{n=1}^{\infty} a_{n} ?\) (Hint: Write \(2 a_{2 n} / a_{n}\) as a product of \(n\) factors each smaller than \(1 /(1+x /(2 n))\)

For the following exercises, indicate whether each of the following statements is true or false. If the statement is false, provide an example in which it is false.Suppose that \(a_{n}\) is a sequence of positive real numbers and that \(\sum_{n=1}^{\infty} a_{n}\) converges. Suppose that \(b_{n}\) is an arbitrary sequence of ones and minus ones. Does \(\sum_{n=1}^{\infty} a_{n} b_{n}\) necessarily converge?

The kth term of each of the following series has a factor \(x^{k}\). Find the range of \(x\) for which the ratio test implies that the series converges. $$ \sum_{k=1}^{\infty} \frac{x^{k}}{k !} $$

Does there exist a number \(p\) such that \(\sum_{n=1}^{\infty} \frac{2^{n}}{n^{p}}\) converges?

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