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Magazine Chapter 9: Problem 32
\(\sum_{n=2}^{\infty}\left(1-\frac{1}{n}\right)^{n}\)
Short Answer
Expert verified
The series diverges.
Step by step solution
01
Identifying the Type of Series
Recognize that the series is \(\sum_{n=2}^{\infty}\left(1-\frac{1}{n}\right)^{n}\) and try to determine its convergence or divergence. Evaluate the function \(a_n = \left(1-\frac{1}{n}\right)^{n}\) as \(n\) approaches infinity.
02
Expressing the Series in Terms of Exponential
Recall the exponential limit \(\left(1-\frac{1}{n}\right)^{n} \approx e^{-1}\) as \(n\rightarrow\infty\) using the approximation \(\left(1+\frac{x}{n}\right)^{n} \approx e^{x}\) for \(x=-1\). Hence, \(a_n\) converges to \(e^{-1}\).
03
Testing Convergence with Limit Comparison Test
Use the limit comparison test with a known convergent or divergent series. Consider the series \(b_n = \frac{1}{n}\). Compute the limit \(\lim_{n\to\infty} \frac{a_n}{b_n} = \lim_{n\to\infty} n\left(1-\frac{1}{n}\right)^n.\) Simplify this expression to see if it leads to a constant.
04
Calculating the Limit
To compute \(\lim_{n\to\infty} n\left(1-\frac{1}{n}\right)^n\), start by using the property \(\left(1-\frac{1}{n}\right)^n = e^{-1} e^{-a_n}\) where the correction term \(e^{-a_n} \approx 1-\frac{1}{n} \). Therefore, \(n\left(1-\frac{1}{n}\right)^n \approx n\cdot\frac{1}{ne}\) converges to \(\frac{1}{e}\).
05
Concluding the Convergence
Since \(\lim_{n\rightarrow\infty} \frac{a_n}{b_n} = \frac{1}{e}\) which is a non-zero constant, \(\sum_{n=2}^{\infty} a_n\) and \(\sum_{n=2}^{\infty} \frac{1}{n}\) either both converge or both diverge. Given that \(\sum_{n=2}^{\infty} \frac{1}{n}\) diverges, the original series \(\sum_{n=2}^{\infty} \left(1-\frac{1}{n}\right)^{n}\) also diverges.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Limit Comparison Test
The limit comparison test is a powerful tool to determine if an infinite series converges or diverges.
When faced with a series \(\sum a_n\), comparing it to another series \(\sum b_n\) that is easier to analyze can simplify the problem. Here's how it works:
When faced with a series \(\sum a_n\), comparing it to another series \(\sum b_n\) that is easier to analyze can simplify the problem. Here's how it works:
- Choose a series \(\sum b_n\) that is similar in form to \(\sum a_n\).
- Ensure you know whether \(\sum b_n\) is convergent or divergent.
- Compute the limit \(L = \lim_{{n \to \infty}} \frac{a_n}{b_n}\).
- If \(L\) is a positive and finite number, and \(L eq 0\), then both \(\sum a_n\) and \(\sum b_n\) will either both converge or both diverge.
Exponential Limit
An exponential limit is a particular calculation in which an expression approaches an exponential function's base value as a variable approaches infinity.
In the given series, we recognize that the general term \(a_n = \left(1-\frac{1}{n}\right)^{n}\) is linked to an exponential limit. The key result used here is:
In the given series, we recognize that the general term \(a_n = \left(1-\frac{1}{n}\right)^{n}\) is linked to an exponential limit. The key result used here is:
- When \(n\) approaches infinity, \(\left(1+\frac{x}{n}\right)^{n}\) approaches \(e^x\).
- Setting \(x = -1\), we find \(\left(1-\frac{1}{n}\right)^n \approx e^{-1}\).
Divergent Series
A divergent series is one in which the sum of its infinite terms does not lead to a finite number.
Unlike convergent series, these continue to increase without bound or oscillate indefinitely.
In this exercise, after performing the limit comparison test, we concluded that the series \(\sum \left(1-\frac{1}{n}\right)^{n}\) diverges. Here's why:
Unlike convergent series, these continue to increase without bound or oscillate indefinitely.
In this exercise, after performing the limit comparison test, we concluded that the series \(\sum \left(1-\frac{1}{n}\right)^{n}\) diverges. Here's why:
- We compared it with the harmonic series \(\sum \frac{1}{n}\), a classic example of a divergent series.
- The comparison provided a non-zero and finite constant result in the limit comparison test.
- Since the harmonic series diverges, and our original series has the same divergence characteristics, it must also diverge.
