Problem 24 - Single Variable Calculus

Question

Determine whether the given series is convergent or divergent. $$ \sum_{n=1}^{\infty} \frac{e^{1 / n}}{n^{2}} $$

Step-by-Step Solution

Answer

The given series \(\sum_{n=1}^{\infty} \frac{e^{1 / n}}{n^{2}}\) is convergent, as determined by the Comparison Test when compared to the convergent p-series \(\sum_{n=1}^{\infty} \frac{1}{n^{2}}\).

Step 1: Identify the Comparison Test and the series to be compared

We will use the Comparison Test and compare the given series \(\sum_{n=1}^{\infty} \frac{e^{1 / n}}{n^{2}}\) to the known convergent p-series \(\sum_{n=1}^{\infty} \frac{1}{n^{2}}\).

Step 2: Perform the comparison

In order for the Comparison Test to be useful, we need to show the terms in our given series are bounded above by those of the convergent p-series. By analyzing the series, we see that \(e^{1 / n} \geq 1\) for all \(n \geq 1\), as the exponential function is always positive. Thus, \(\frac{e^{1 / n}}{n^{2}} \geq \frac{1}{n^{2}}\) Since \(e^{1 / n} \geq 1\), it follows that \(e^{1 / n} \leq e\), as \(n\) approaches infinity. Therefore, \(\frac{e^{1 / n}}{n^{2}} \leq \frac{e}{n^{2}}\) We have \(\frac{1}{n^{2}} \leq \frac{e^{1 / n}}{n^{2}} \leq \frac{e}{n^{2}}\) According to the Comparison Test, if the terms of a series have all the values positive and bounded above by those of a convergent series, then it is convergent. Conversely, if the terms of a series have all the values positive and bounded below by those of a divergent series, then it is divergent.

Step 3: Apply the Comparison Test

Since the series \(\sum_{n=1}^{\infty} \frac{1}{n^{2}}\) is a convergent p-series, by the Comparison Test, the given series, \(\sum_{n=1}^{\infty} \frac{e^{1 / n}}{n^{2}}\), is also convergent.

Key Concepts

Comparison Test

Understanding the Comparison Test can be incredibly useful when you're tackling infinite series. This test comes in handy especially when you are trying to determine if a series converges or not. Essentially, the Comparison Test compares two series to draw conclusions about their convergence or divergence.

Here's the gist: if you have a series with all positive terms, and you can find another series with larger positive terms that converges, then your original series converges as well. On the flip side, if the terms of your series are always larger than the terms of a divergent series, then your series must diverge too.

Key Criteria for the Comparison Test

For this test to be valid, the series you're comparing must meet two conditions:

• The series must have positive terms.
• The series you compare to must be a known convergent or divergent series.

If your series meets these criteria, you can make a solid comparison to determine its convergence or divergence. Remember to always use a series whose convergence is already known to you. This will serve as a benchmark, allowing you to deduce the behavior of the series in question.

p-series

A p-series is an infinite series that has a very particular form. It looks like this: \[\sum_{n=1}^{\infty} \frac{1}{n^p}\] where 'p' is a constant power. The convergence or divergence of a p-series directly depends on the value of this 'p'.

Here's the rule:

• If 'p' is greater than 1, the series converges.
• If 'p' is less than or equal to 1, the series diverges.

This criterion makes p-series a valuable tool, since once you determine the value of 'p', you instantly know if the series converges or diverges, without doing any additional calculations.

For example, the series \(\sum_{n=1}^{\infty} \frac{1}{n^{2}}\) is known to converge because here 'p' equals 2, which is greater than 1. This series serves as a reference when using comparison tests, as its behavior is a known benchmark in the study of infinite series.

Infinite Series

An infinite series is basically a sum that keeps going on forever. It's made up of an infinite list of terms added together. While this concept might seem complex, it's just an extension of the addition you've learned since grade school – but with no clear end in sight!

While some infinite series grow larger and larger with each added term, and therefore 'diverge', others settle into a fixed sum, hence 'converge'. The convergence or divergence of an infinite series isn’t about what happens at the start, but rather what happens as you move towards infinitely many terms.

Therefore, inquiring about the convergence of the series \(\sum_{n=1}^{\infty} \frac{e^{1 / n}}{n^{2}}\) is really asking about its behavior as 'n' becomes very large. Because the term \(e^{1/n}\) becomes closer to 1 as 'n' increases, the series resembles the convergent p-series \(\sum_{n=1}^{\infty} \frac{1}{n^{2}}\), which is why we used the comparison test to determine its convergence.