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Chapter 2: Problem 1

Suppose the kth partial sum of \(\sum_{n=1} x_{n}\) is \(s_{k}=\frac{k}{k+1} .\) Find the series, that is find \(x_{n}\), prove that the series converges, and then find the limit.

Short Answer

Expert verified
The series is given by \(x_n = \frac{2n-1}{n(n+1)}\) and converges to the limit 1.

Step by step solution

01

Differentiate Partial Sum Expression

Given that the kth partial sum of the series is \(s_k = \frac{k}{k+1}\). To find the terms of the series \(x_n\), recall that the partial sum \(s_k\) is the sum of the first \(k\) terms. The nth term is given by \(x_n = s_n - s_{n-1}\).
02

Find Expression for nth Term

To find \(x_n\), calculate: \[ x_n = s_n - s_{n-1} = \frac{n}{n+1} - \frac{n-1}{n} = \frac{n^2 - (n^2 - 2n + 1)}{n(n+1)} = \frac{2n - 1}{n(n+1)}. \] Thus, \(x_n = \frac{2n-1}{n(n+1)}\).
03

Convergence Analysis

To prove the series converges, use the fact that a series \(\sum x_n\) converges if its terms tend to 0 as \(n\to\infty\). For \(x_n = \frac{2n-1}{n(n+1)}\), as \(n\to\infty\), \(x_n\) behaves like \(\frac{2n}{n^2}\). Hence, \(x_n \to 0\). Additionally, use the telescoping nature of the sequence; \(\sum x_n = s_n \to 1\) as \(n\to\infty\).
04

Find the Limit of the Overall Sum

Given \(s_k = \frac{k}{k+1}\), as \(n\to\infty\), \(s_n\to 1\). This means the series converges to 1.

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Key Concepts

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

Partial Sum
In a series, the partial sum represents the sum of a sequence's first few terms. It鈥檚 denoted as \(s_k = \frac{k}{k+1}\), reflecting the accumulated sum up to the \(k\)-th term. The idea is simple but powerful: by understanding the pattern of these sums, we can infer a lot about the series. For students grappling with the concept, it's essential to grasp that the partial sum \(s_k\) gives us a sneak peek into how the series behaves as more terms are added.
  • For example, if \(s_k\) appears to stabilize as \(k\) increases, this indicates a potential numerical limit.
  • In this exercise, \(s_k\) chases closer to 1 as \(k\) increases, providing insight for further analysis.
Understanding this offers a gateway to exploring whether the whole series converges or diverges.
Telescoping Series
A telescoping series is a special type of series where most terms cancel out, like the collapsing segments of a telescope. In these series, determining the individual terms helps unlock the key pattern. Here, the term \(x_n\) is discovered through the relation:\[ x_n = s_n - s_{n-1} \]This formula gives us insight into how consecutive sums overlap, making it dramatically simpler to find the sequence's overall behavior.
  • In our example, calculating \(x_n = \frac{2n-1}{n(n+1)}\) showcases the typical telescoping effect.
  • As \(n\) grows, the precise canceling of terms helps us visualize its behavior as \(n\) approaches infinity.
Telescoping reveals not just the terms but also sets the stage for convergence analysis.
Series Limit
The limit of a series is what we can expect the series to equal when all its terms are summed indefinitely. To determine this, we observe the partial sums; here, \(s_k = \frac{k}{k+1}\). As \(n\) becomes very large, \(s_n\) approaches the limit. For this exercise, \(s_n \overset{n\to\infty}{\longrightarrow} 1\). This result tells us that if you summed up all terms from the beginning to infinity, the sum would stabilize close to this number.
  • Recognizing this limit is pivotal. It impacts mathematics ranging from analysis to applied fields like physics or economics, signifying the ultimate value of infinitesimal accumulations.
  • Once students see this pattern, identifying limits in similar functions becomes considerably easier.
Exploring limits through exercises like these nurtures a deeper understanding of infinite processes.
Term Calculation
Finding the nth term of a series, \(x_n\), involves a bit of detective work where the partial sums play a crucial role. This is because each term is essentially the difference between two consecutive partial sums: \[ x_n = s_n - s_{n-1} \]In our example, this calculation is completed by resolving \[ x_n = \frac{2n-1}{n(n+1)} \]This formula reveals how complex terms arise from simple arithmetic manipulations of partial sums.
  • Solving for individual terms like this not only broadens your mathematical toolkit but also fortifies your problem-solving skills.
  • The simplicity behind the computation method hides the profound complexity鈥攁nd beauty鈥攐f infinite series.
Accurate term calculation is foundational for deeper inquiry into convergence and series behavior, offering practical insights into the architecture of reasons.

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Most popular questions from this chapter

Suppose \(\left\\{x_{n}\right\\}\) is a sequence, and \(a_{n}:=\sup \left\\{x_{k}: k \geq n\right\\}\) and \(b_{n}:=\sup \left\\{x_{k}: k \geq n\right\\}\) as before. a) Prove that if \(a_{\ell}=\infty\) for some \(\ell \in \mathbb{N},\) then \(\limsup x_{n}=\infty\). b) Prove that if \(b_{\ell}=-\infty\) for some \(\ell \in \mathbb{N},\) then \(\liminf x_{n}=-\infty\).

a) Show that the alternating harmonic series \(\sum \frac{(-1)^{n-1}}{n}\) has a rearrangement such that for any \(x

(Easy): Prove the following stronger version of Lemma 2.2.12, the ratio test. Suppose \(\left\\{x_{n}\right\\}\) is a sequence such that \(x_{n} \neq 0\) for all \(n\). a) Prove that if there exists an \(r<1\) and \(M \in \mathbb{N}\) such that for all \(n \geq M\) we have $$ \frac{\left|x_{n+1}\right|}{\left|x_{n}\right|} \leq r $$ then \(\left\\{x_{n}\right\\}\) converges to \(0 .\) b) Prove that if there exists an \(r>1\) and \(M \in \mathbb{N}\) such that for all \(n \geq M\) we have $$ \frac{\left|x_{n+1}\right|}{\left|x_{n}\right|} \geq r $$ then \(\left\\{x_{n}\right\\}\) is unbounded.

In the following exercises, feel free to use what you know from calculus to find the limit, if it exists. But you must prove that you found the correct limit, or prove that the series is divergent. Is the sequence \(\left\\{\frac{n}{n^{2}+1}\right\\}\) convergent? If so, what is the limit?

Let \(\left\\{x_{n}\right\\}\) be a sequence defined by $$x_{n}:=\left\\{\begin{array}{ll} n & \text { if } n \text { is odd } \\ 1 / n & \text { if } n \text { is even } \end{array}\right.$$ a) Is the sequence bounded? (prove or disprove) b) Is there a convergent subsequence? If so, find it.
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