Problem 11 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

Determine whether the given series converges or diverges. $$ \sum_{n=1}^{\infty} \ln \left(2+\frac{1}{n}\right) $$

Step-by-Step Solution

Answer

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

Step 1: Analyze the series term

First, observe the terms of the series \(\text{ln} \bigg(2 + \frac{1}{n}\bigg)\). As \(n\) tends to infinity, \(\frac{1}{n}\) approaches 0, so each term gets closer to \(\text{ln}(2)\).

Step 2: Simplify the logarithmic term

To simplify the series term \(\text{ln} \bigg(2 + \frac{1}{n}\bigg)\), let's expand \(\text{ln} \bigg(2 + \frac{1}{n}\bigg)\) using the series expansion for logarithms around small \(\frac{1}{n}\). This gives us approximately \(\text{ln}(2) + \text{ln} \bigg(1 + \frac{1}{2n} \bigg)\).

Step 3: Apply the series expansion

Utilize the series expansion \(\text{ln} (1+x) \sim x\) for small \(x\), such that \(\text{ln} \bigg(1 + \frac{1}{2n} \bigg) \approx \frac{1}{2n}\). Thus, \(\text{ln} \bigg(2 + \frac{1}{n}\bigg) \approx \text{ln}(2) + \frac{1}{2n}\).

Step 4: Separate the series

Rewrite the series as follows: \(\text{ln}(2) + \sum_{n=1}^{\infty} \frac{1}{2n}\). The term \(\text{ln}(2)\) is a constant and thus does not affect the convergence of the series itself. Focus on the series \(\sum_{n=1}^{\infty} \frac{1}{2n}\).

Step 5: Test convergence of simplified series

We know that \(\sum_{n=1}^{\infty} \frac{1}{n}\) is the harmonic series and diverges. Therefore, \(\sum_{n=1}^{\infty} \frac{1}{2n}\) is simply \(\frac{1}{2}\) times the harmonic series, which still diverges.

Step 6: Conclusion

Since the series \(\sum_{n=1}^{\infty} \frac{1}{2n}\) diverges, the original series \(\sum_{n=1}^{\infty} \text{ln} \bigg(2 + \frac{1}{n}\bigg)\) also diverges.

Key Concepts

logarithmic series

The original exercise focuses on the series \(\text{ln} \bigg(2 + \frac{1}{n}\bigg)\). At the core of this problem is the concept of a **logarithmic series**, which involves the natural logarithm function log\((x)\) applied to the terms of a sequence. In this case, we need to determine if the series \(\text{ln} \bigg(2 + \frac{1}{n}\bigg)\) converges or diverges as n approaches infinity.
To analyze such series, we often use logarithmic properties and approximations. For small values of x, the natural logarithm has the series expansion \(\text{ln}(1 + x) \approx\ x\). We use this approximation to simplify the given series term \( \text{ln}\bigg(2 + \frac{1}{n}\bigg)\). As n tends to infinity, \(\frac{1}{n}\) becomes very small, and thus our term can be written as \(\text{ln}(2) + \text{ln}\bigg(1 + \frac{1}{2n}\bigg)\), allowing us to further simplify using the approximation for \(\text{ln}(1 + x)\).
Understanding the behavior and properties of the logarithmic function at different ranges can be very helpful in simplifying and analyzing series.

harmonic series

The **harmonic series** is a fundamental concept in understanding series convergence and divergence. The harmonic series is given by: \[ \sum_{n=1}^{\text{∞}} \frac{1}{n} \] This series is known to diverge, meaning that its sum grows without bound as n increases.
In our original problem, we derived that the original series transforms into \( \text{ln}(2) + \frac{1}{2n} \). The term \( \text{ln}(2) \) is a constant and doesn't affect the convergence. Hence, we focus on \( \sum_{n=1}^{\text{∞}} \frac{1}{2n} \), which is essentially half the harmonic series.
Since the harmonic series itself diverges, any multiple of it, including the one with \(\frac{1}{2}\), also diverges. This conclusion helps us determine the behavior of complex series by comparing them to well-known series like the harmonic series.

series divergence

The concept of **series divergence** is crucial in our problem. When we say a series diverges, it means the sum of its terms keeps growing without bound. This happens if the terms of the series do not approach zero fast enough.
In the step-by-step solution, we concluded that the series \( \sum_{n=1}^{\text{∞}} \frac{1}{2n} \) diverges because it is simply a scaled version of the harmonic series.
Divergence can be tested using various methods, such as comparing to known divergent series (like we did here), the n-th term test (if the limit of the n-th term as n approaches infinity is not zero, the series diverges), or using the Ratio Test and other convergence tests. Understanding that the harmonic series diverges is a foundational tool for analyzing other series.

series expansion

The **series expansion** of functions, particularly for simplifying terms within a series, is a very powerful analytical tool. In our exercise, the concept of the logarithmic series expansion was leveraged to simplify the term \( \text{ln}\bigg(2 + \frac{1}{n}\bigg) \).
Specifically, we used the logarithmic series expansion: \( \text{ln}(1 + x) \approx\ x \), valid for small x, to approximate \( \text{ln}\bigg(2 + \frac{1}{n}\bigg) \approx\ \text{ln}(2) + \frac{1}{2n} \).
This approximation allowed us to express the original series in a more manageable form, eventually highlighting its resemblance to the harmonic series. Successful series expansion can simplify complex terms and make the analysis of convergence/divergence more straightforward.
Grasping series expansion is essential in calculus and beyond, as it can simplify solving problems across various mathematical fields.