Question

In the following exercises, use an appropriate test to determine whether the series converges. $$a_n={(1+\frac{1}{n^2})}^n(\text{ Hint }{(1+\frac{1}{n^2})}^{n^2}\approx e.)$$

Short answer

The series diverges because the terms approach 1, not zero.

Step-by-step solution

Understand the Series

The term in question is $a_n={(1+\frac{1}{n^2})}^n$. We are tasked with determining if the series of $a_n$ converges. The hint suggests that ${(1+\frac{1}{n^2})}^{n^2}\approx e$.

Transformation using Exponent Properties

Consider the hint, which is similar to the limit definition of $e$, ${(1+\frac{1}{n})}^n\to e$ as $n\to \mathrm{\infty }$. For ${(1+\frac{1}{n^2})}^n$, we can write this as $\text{Missing left or extra right}$.

Apply the Limit to Transform Expression

We know from the hint that ${(1+\frac{1}{n^2})}^{n^2}\approx e$. If ${(1+\frac{1}{n^2})}^{n^2}\to e$ as $n\to \mathrm{\infty }$, the expression ${\left(e\right)}^{1/n}\to 1$ as $n\to \mathrm{\infty }$.

Convergence Test Using Limits

Since $$\underset{n\to \mathrm{\infty }}{lim}{a}_n=\underset{n\to \mathrm{\infty }}{lim}{\left({(1+\frac{1}{n^2})}^{n^2}\right)}^{1/n}=1$$, for the series to converge, individual terms should approach zero, which does not happen here.

Conclusion: Determine the Nature of the Series

Since the terms $a_n\to 1$ as $n\to \mathrm{\infty }$ instead of zero, the series diverges based on the divergence test.

Key concepts

Divergence Test

The Divergence Test, also known as the nth-term test, is a simple but powerful tool in determining the convergence of a series. Here's how it works: if the limit of the terms of a series as the index goes to infinity doesn't equal zero, then the series must diverge. It's crucial because many series can only converge if their terms get smaller and closer to zero as they progress.

For example, in the series given by terms $a_n={(1+\frac{1}{n^2})}^n$, we need to use the divergence test to check if $a_n$ converges towards zero.

• By transforming the expression as indicated, we find that $\underset{n\to \mathrm{\infty }}{lim}{a}_n=1$.
• Since the terms do not approach zero, the divergence test conclusively indicates that the series diverges.

In this way, the divergence test helps quickly rule out convergence when series terms remain non-zero.

Limit Comparison Test

The Limit Comparison Test is a useful method for determining the convergence of series, especially when the form of the series resembles another, simpler series with known behavior. The idea is to compare the terms of our series with the terms of a benchmark series.

This test states that if we have two series $\sum a_n$ and $\sum b_n$, and $\underset{n\to \mathrm{\infty }}{lim}\frac{a_n}{b_n}=c$ where $c$ is a positive finite number, then either both series converge or both diverge.

• The test assumes that $b_n$ is a straightforward series whose convergence is known, such as a p-series $1/n^p$.
• This is particularly beneficial for series that are not evidently geometric or simple to analyze on their own.

Although this test wasn't used on our original series, it offers a robust alternative for comparison-based determination of convergence.

Series Convergence

Series convergence is a fundamental concept in calculus and analysis that determines whether the sum of the infinite terms of a sequence results in a finite number. If a series converges, its terms approach a specific sum as the number of terms increases indefinitely.

To decide convergence, mathematicians use different tests, like the divergence test or limit comparison test, depending on the structure of the series.

• A convergent series implies that as you add more terms, the cumulative value will stabilize at a particular point.
• Many series fail to satisfy this criterion, indicating divergence, as observed in the given exercise where the term $a_n$ approaches 1.

Understanding series convergence is crucial because convergent series can have significant applications in mathematical modeling and problem-solving, providing solutions that are exact and well-defined.