Problem 54 - Essential Calculus

Question

Determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty} \frac{n+1}{2 n-1} $$

Step-by-Step Solution

Answer

The series \( \sum_{n=1}^{\infty} \frac{n+1}{2n-1} \) diverges.

Step 1: Identify a suitable comparison series

Choose the series \( \sum_{n=1}^{\infty} \frac{1}{n} \) as comparison series (harmonic series). For the harmonic series, it's well-known that it diverges.

Step 2: Use the comparison test

Start the comparison. We have: \( \frac{n+1}{2n-1} \) ≥ \( \frac{1}{2n} \) for all n. This can be rearranged to: \( n+1 \) ≥ 1. This inequality holds for all n ≥ 1, hence the comparison test can be applied.

Step 3: Apply the comparison test

Since the comparison series \( \sum_{n=1}^{\infty} \frac{1}{n} \) is divergent and our series \( \sum_{n=1}^{\infty} \frac{n+1}{2n-1} \) is greater than or equal to the comparison series, we conclude using the comparison test that our series also diverges.

Key Concepts

Comparison Test

The comparison test is a useful tool when determining whether an infinite series converges or diverges. In this test, we compare the series in question with another series, whose convergence properties we already know. Here's how it works:
When considering a series \(\sum a_n\), select another series \(\sum b_n\). This second series should be one that is known to converge or diverge. Then, look at the terms of the series:

• If \(0 \leq a_n \leq b_n\) for all terms, and \(\sum b_n\) converges, then \(\sum a_n\) also converges.
• Conversely, if \(a_n \geq b_n \geq 0\) for all terms, and \(\sum b_n\) diverges, then \(\sum a_n\) also diverges.

In the original exercise, the series \(\sum_{n=1}^{\infty} \frac{n+1}{2n-1}\) was compared with the harmonic series \(\sum_{n=1}^{\infty} \frac{1}{n}\), known to diverge. Since the terms \(\frac{n+1}{2n-1}\) are individually equal to or greater than the terms of \(\frac{1}{n}\), the comparison test concludes that the original series must also diverge.

Harmonic Series

The harmonic series is one of the most important series in mathematics, famous for its simplicity and fascinating property of divergence. It is defined as the sum of reciprocals of natural numbers:
\[\sum_{n=1}^{\infty} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots\]
Despite the terms decreasing and approaching zero, the harmonic series diverges. This means that if you keep adding terms indefinitely, the total will grow without bound. This characteristic of the harmonic series makes it a perfect benchmark for applying the comparison test, especially when assessing series with terms that behave similarly to \(\frac{1}{n}\).
In the exercise, the harmonic series is used as the comparison benchmark because it's well known to diverge, setting a clear precedent for comparison with other series.

Divergence

Divergence in the context of infinite series refers to the property where a series does not settle to a finite limit. If as you add more terms in a series and the sum keeps growing without any limit, the series is said to diverge.
In more formal terms, a series \(\sum a_n\) diverges if its sequence of partial sums does not have any finite limit. Divergence is opposite to convergence, where the series approaches a specific value.
In the given exercise, the divergence was shown using the comparison test. By proving that the series \(\sum_{n=1}^{\infty} \frac{n+1}{2n-1}\) is greater than the divergent harmonic series, it follows that it too diverges. Recognizing divergence is crucial in determining the behavior of infinite series, which can have significant implications in areas like analysis and number theory.