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Chapter 4: Problem 10

In the following exercises, compute the general term an of the series with the given partial sum Sn. If the sequence of partial sums converges, find its limit S. $$ S_{n}=\frac{n(n+1)}{2}, n \geq 1 $$

Short Answer

Expert verified
The general term is \( a_n = n \), and the series diverges.

Step by step solution

01

Identify the Relationship Between Partial Sum and General Term

To find the general term \( a_n \) of the series when given the partial sum \( S_n \), use the relationship: \( a_n = S_n - S_{n-1} \).
02

Calculate S_n - S_{n-1}

Given \( S_n = \frac{n(n+1)}{2} \), find \( S_{n-1} \) by substituting \( n-1 \) into the formula: \( S_{n-1} = \frac{(n-1)n}{2} \). Now compute \( a_n = S_n - S_{n-1} = \frac{n(n+1)}{2} - \frac{(n-1)n}{2} \).
03

Simplify the Expression for a_n

Simplify the expression \( a_n = \frac{n(n+1)}{2} - \frac{(n-1)n}{2} \) by distributing and combining like terms: \( a_n = \frac{n^2 + n - n^2 + n}{2} = \frac{2n}{2} = n \). Thus, the general term \( a_n = n \).
04

Determine the Convergence of the Sequence of Partial Sums

Since \( a_n = n \) and the terms increase without bound, the partial sums \( S_n \) do not converge to a finite limit. Thus, the sequence of partial sums diverges.

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Key Concepts

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

Partial Sum
The concept of partial sums is an essential building block in understanding infinite series. When you have a series, it's essentially a list of numbers that you're adding together. The sum doesn't happen all at once but builds over time. This is where partial sums come in.
To illustrate:
  • Suppose you have an infinite series, which is a list of numbers added consecutively.
  • The partial sum is the sum of the first few terms of this series.
  • As you keep adding more terms, the partial sum sequence grows.
In the given exercise, the partial sum is represented as \( S_n = \frac{n(n+1)}{2} \). This formula calculates the sum of the first \( n \) terms. A key step is understanding how this formula evolves as \( n \) changes. It helps us understand how the entire series behaves as it extends.
General Term
Finding the general term of a series is like finding the recipe for each new dish in an endless course meal. It tells you exactly what you're adding to the bowl at each step. The series described in the exercise originally gives us every partial sum. But to understand and work with it, we need to unravel it a bit.Here's how to find the general term from the partial sum:
  • We use the formula \( a_n = S_n - S_{n-1} \) to get the general recipe or term for each step in the series.
  • By substituting and simplifying, \( a_n = \frac{n(n+1)}{2} - \frac{(n-1)n}{2} \), you derive that each term added is \( n \).
This step is crucial as it changes our perspective from looking at accumulated sums to understanding the individual contributions at each stage.
Divergence
Divergence is a significant concept when analyzing series. When we say a series diverges, we mean its partial sums grow without bound, never settling at a particular value. It means that the series doesn't have a finite sum.
For a series to converge, the sequence of its partial sums should approach a specific limit. However, divergence means no such boundary exists.
  • In our exercise, since \( a_n = n \), which implies that every term added gets larger as \( n \) increases, the partial sums increase indefinitely.
  • The lack of a plateau in the partial sums \( S_n \) indicates divergence.
Understanding divergence helps us identify when a series doesn't "behave" in a way that leads to a specific, calculable sum as \( n \) heads to infinity.

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Most popular questions from this chapter

The following series do not satisfy the hypotheses of the alternating series test as stated. In each case, state which hypothesis is not satisfied. State whether the series converges absolutely.Sometimes the alternating series \(\sum_{n=1}^{\infty}(-1)^{n-1} b_{n}\) converges to a certain fraction of an absolutely convergent series \(\sum_{n=1}^{\infty} b_{n}\) a faster rate. Given that \(\sum_{n=1}^{\infty} \frac{1}{n^{2}}=\frac{\pi^{2}}{6}\), find \(S=1-\frac{1}{2^{2}}+\frac{1}{3^{2}}-\frac{1}{4^{2}}+\cdots .\) Which of the series \(6 \sum_{n=1}^{\infty} \frac{1}{n^{2}}\) and \(S \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{2}}\) gives a better estimation of \(\pi^{2}\) using 1000 terms?

Use the ratio test to determine whether \(\sum_{n=1}^{\infty} a_{n}\) converges, where \(a_{n}\) is given in the following problems. State if the ratio test is inconclusive. $$ \sum_{n=1}^{\infty} \frac{\left(2^{n} n !\right)^{2}}{(2 n)^{2 n}} $$

Is the series convergent or divergent? $$ \sum_{n=1}^{\infty} \ln \left(\frac{n+1}{n}\right) $$

The following series do not satisfy the hypotheses of the alternating series test as stated. In each case, state which hypothesis is not satisfied. State whether the series converges absolutely.Show that the alternating series \(\frac{2}{3}-\frac{3}{5}+\frac{4}{7}-\frac{5}{9}+\cdots\) does not converge. What hypothesis of the alternating series test is not met?

In the following exercises, use either the ratio test or the root test as appropriate to determine whether the series \(\sum a_{k}\) with given terms \(a_{k}\) converges, or state if the test is inconclusive. $$ a_{n}=\left(1-\frac{1}{n}\right)^{n^{2}} $$
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