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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. $$ S_{n}=1-\frac{1}{n}, n \geq 2 $$

Short Answer

Expert verified
The general term is \( a_n = \frac{1}{n(n-1)} \), and the limit of the partial sums is 1.

Step by step solution

01

Understand the Problem

We are given the partial sum of a series as \( S_n = 1 - \frac{1}{n} \). Our goal is to find the general term \( a_n \) for the series and, if it converges, find the limit of the sequence of partial sums (the sum of the series).
02

Identify the Relationship

We know that the general term \( a_n \) of the series is the difference between consecutive terms of the partial sums: \[ a_n = S_n - S_{n-1} \].
03

Calculate \( a_n \)

Substitute the given \( S_n \) into the equation: \[ a_n = \left(1 - \frac{1}{n}\right) - \left(1 - \frac{1}{n-1}\right) \]. Simplify to get: \[ a_n = \frac{1}{n-1} - \frac{1}{n} \].
04

Simplify \( a_n \)

Recognize this as a difference of fractions: \[ a_n = \frac{n - (n-1)}{n(n-1)} = \frac{1}{n(n-1)} \].
05

Check for Convergence

The sequence of partial sums \( S_n = 1 - \frac{1}{n} \) converges as \( n \to \infty \). Observe that: \[ \lim_{n \to \infty} S_n = 1 \].
06

Conclusion

The general term \( a_n \) is \( \frac{1}{n(n-1)} \) and the limit of the partial sums is 1.

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Key Concepts

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

Partial Sum
A partial sum is the sum of the first "n" terms of a series. It is denoted as \( S_n \). For instance, when you're trying to find the 5th partial sum, that means you sum up the first 5 terms of the series. This concept is important because it helps us understand whether a series converges or not by looking at how these sums behave as \( n \) increases.
In this exercise, the partial sum given is \( S_n = 1 - \frac{1}{n} \). Each \( S_n \) provides insight into the nature of the series as \( n \) becomes very large. As \( n \) approaches infinity, our sequence of partial sums tends to get closer to a specific value.
General Term of a Series
The general term of a series, represented by \( a_n \), is essentially the formula that describes the nth term of a sequence. It allows us to understand what each individual component of the series looks like.
We find the general term by observing that \( a_n = S_n - S_{n-1} \). In this exercise, using the partial sum \( S_n \), we calculated that \( a_n = \frac{1}{n-1} - \frac{1}{n} \). Simplifying this gives \( a_n = \frac{1}{n(n-1)} \). The general term clarifies how each term behaves or changes within the series, which is crucial for understanding the series' overall behavior.
Limit of a Sequence
The concept of the limit of a sequence is about determining the value that the terms of a sequence approach as the index \( n \) becomes very large. When we talk about convergence, we mean whether these terms settle towards a fixed number.
In this exercise, we noticed that the sequence of partial sums \( S_n = 1 - \frac{1}{n} \) converges to 1 as \( n \to \infty \). This means that as we take more terms, the partial sums get closer and closer to 1. Knowing the limit is vital because it indicates the behavior of an infinite series and whether it tends to a finite, stable number.
Sequence of Partial Sums
A sequence of partial sums is simply a sequence created by summing the terms of a series up to a certain point, and then observing these sums as a series of its own. It's like having a checklist where you note down the total you have so far, which keeps getting updated with each new term you add.
In this case, our sequence of partial sums \( S_n \) is \( 1 - \frac{1}{n} \). Observing this sequence helps us identify the convergence of the series: as \( n \) increases, the sequence of partial sums converges to the number 1. This means the entire series approaches that same finite number in the long run.

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Most popular questions from this chapter

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 \(1 /(1+x /(2 n))\)

Use the root test to determine whether \(\sum_{m=1}^{\infty} a_{n}\) converges, where \(a_{n}\) is as follows. $$ a_{k}=\left(\frac{k-1}{2 k+3}\right)^{k} $$

[T] Find the first 1000 digits of \(\pi\) using either a computer program or Internet resource. Create a bit sequence \(b_{n}\) by letting \(b_{n}=1\) if the \(n\) th digit of \(\pi\) is odd and \(b_{n}=0\) if the \(n\) th digit of \(\pi\) is even. Compute the average value of \(b_{n}\) and the average value of \(d_{n}=\left|b_{n+1}-b_{n}\right|, n=1, \ldots, 999 .\) Does the sequence \(b_{n}\) appear random? Do the differences between successive elements of \(b_{n}\) appear random?

Is the sequence bounded, monotone, and convergent or divergent? If it is convergent, find the limit. $$ a_{n}=\ln \left(\frac{1}{n}\right) $$

Use the root test to determine whether \(\sum_{m=1}^{\infty} a_{n}\) converges, where \(a_{n}\) is as follows. $$ a_{n}=\frac{(\ln (1+\ln n))^{n}}{(\ln n)^{n}} $$

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