Problem 60
Question
Which series in Exercises \(49-68\) converge, and which diverge? Give reasons for your answers. If a series converges, find its sum. $$ \sum_{n=1}^{\infty}\left(1-\frac{1}{n}\right)^{n} $$
Step-by-Step Solution
Verified Answer
The series diverges by the divergence test.
1Step 1: Observing the General Term
Let's consider the general term of the series: \(a_n = \left(1 - \frac{1}{n}\right)^n\). We need to understand the behavior of \(a_n\) as \(n\) approaches infinity.
2Step 2: Evaluating the Limit of the General Term
We know that \(\left(1 + \frac{x}{n}\right)^n \approx e^x\) as \(n\) approaches infinity. By analyzing \(\left(1 - \frac{1}{n}\right)^n\), we find that it approaches \(e^{-1}\) as \(n\) goes to infinity. Thus, the terms have a finite limit.
3Step 3: Applying the Divergence Test
According to the divergence test (or test for divergence), if \(\lim_{n \to \infty} a_n eq 0\), then the series \(\sum_{n=1}^{\infty} a_n\) diverges. Here, \(\lim_{n \to \infty} \left(1 - \frac{1}{n}\right)^n = \frac{1}{e} eq 0\).
4Step 4: Conclusion on Convergence or Divergence
Since the limit of the terms \(a_n\) approaches a non-zero number, the series diverges by the divergence test. Therefore, there is no finite sum.
Key Concepts
Limit of a SequenceDivergence TestInfinite Series
Limit of a Sequence
Understanding the limit of a sequence is crucial in analyzing series. A sequence is essentially an ordered list of numbers, and the limit is the value that the numbers approach as the index (often denoted as \(n\)) increases. In our exercise, we look at the sequence given by \(a_n = \left(1 - \frac{1}{n}\right)^n\). As \(n\) becomes very large, \(n\) tends to infinity. This is a typical case in calculus where evaluating limits helps disclose the behavior of sequences.
To find this specific limit, we can use known approximations related to the mathematical constant \(e\). Specifically, \(\left(1 + \frac{x}{n}\right)^n\) approaches \(e^x\) as \(n\) approaches infinity. Applying this to \(1 - \frac{1}{n}\) gives us \(e^{-1}\). Therefore, as \(n\) gets larger, the sequence \(\left(1 - \frac{1}{n}\right)^n\) approaches \(\frac{1}{e}\), which is approximately 0.3679.
This behavior is useful to determine whether a series converges or diverges.
To find this specific limit, we can use known approximations related to the mathematical constant \(e\). Specifically, \(\left(1 + \frac{x}{n}\right)^n\) approaches \(e^x\) as \(n\) approaches infinity. Applying this to \(1 - \frac{1}{n}\) gives us \(e^{-1}\). Therefore, as \(n\) gets larger, the sequence \(\left(1 - \frac{1}{n}\right)^n\) approaches \(\frac{1}{e}\), which is approximately 0.3679.
This behavior is useful to determine whether a series converges or diverges.
Divergence Test
The divergence test, also known as the test for divergence, is a simple yet fundamental rule in determining the convergence of an infinite series. It stipulates that if the limit of the sequence \(a_n\) does not equal zero, i.e., \(\lim_{n \to \infty} a_n eq 0\), then the series \(\sum_{n=1}^{\infty} a_n\) definitely diverges.
This criterion is a quick way to identify divergent series. However, passing the divergence test (where the limit is zero) doesn't imply convergence; the series may still diverge. It only helps to identify divergence. In our exercise, when we evaluate \(\lim_{n \to \infty} \left(1 - \frac{1}{n}\right)^n\), we found it is \(\frac{1}{e} eq 0\).
As per the divergence test, the series \(\sum_{n=1}^{\infty} \left(1 - \frac{1}{n}\right)^n\) diverges. Because the terms of the sequence do not trend toward zero, the terms do not sum to a finite number and thus do not converge.
This criterion is a quick way to identify divergent series. However, passing the divergence test (where the limit is zero) doesn't imply convergence; the series may still diverge. It only helps to identify divergence. In our exercise, when we evaluate \(\lim_{n \to \infty} \left(1 - \frac{1}{n}\right)^n\), we found it is \(\frac{1}{e} eq 0\).
As per the divergence test, the series \(\sum_{n=1}^{\infty} \left(1 - \frac{1}{n}\right)^n\) diverges. Because the terms of the sequence do not trend toward zero, the terms do not sum to a finite number and thus do not converge.
Infinite Series
An infinite series is essentially an endless sum of terms from a sequence. Specifically, a series \(\sum_{n=1}^{\infty} a_n\) is the summation of its sequence's terms. To fathom an infinite series, one needs to assess whether it converges (reaches a finite sum) or diverges (does not settle to any particular value).
A series converges when the addition of infinitely many numbers results in a finite sum, which might seem counterintuitive but happens often in calculus and analysis. If the sequence of the partial sums of a series converges to a limit, the series itself converges. There are many tests, like the ratio test, root test, comparison test, and more to ascertain this behavior.
However, if the sum keeps increasing without bound or doesn't settle at a single value, it diverges. In the example exercised, the series \(\sum_{n=1}^{\infty} \left(1 - \frac{1}{n}\right)^n\) diverges, discerned through the divergence test since the individual sequence terms don't approach zero.
A series converges when the addition of infinitely many numbers results in a finite sum, which might seem counterintuitive but happens often in calculus and analysis. If the sequence of the partial sums of a series converges to a limit, the series itself converges. There are many tests, like the ratio test, root test, comparison test, and more to ascertain this behavior.
However, if the sum keeps increasing without bound or doesn't settle at a single value, it diverges. In the example exercised, the series \(\sum_{n=1}^{\infty} \left(1 - \frac{1}{n}\right)^n\) diverges, discerned through the divergence test since the individual sequence terms don't approach zero.
Other exercises in this chapter
Problem 60
Convergence or Divergence Which of the series in Exercises \(55-62\) converge, and which diverge? Give reasons for your answers. $$\sum_{n=1}^{\infty} \frac{n^{
View solution Problem 60
Suppose that \(a_{n}>0\) and \(\lim _{n \rightarrow \infty} n^{2} a_{n}=0 .\) Prove that \(\sum a_{n}\) converges.
View solution Problem 61
Obtain the Taylor series for 1\(/(1+x)^{2}\) from the series for \(-1 /(1+x) .\)
View solution Problem 61
Show that \(\sum_{n=2}^{\infty}\left((\ln n)^{q} / n^{p}\right)\) converges for \(-\infty 1.\) (Hint: Limit Comparison with \(\sum_{n=2}^{\infty} 1 / n^{r}\) fo
View solution