Question

Use the test of your choice to determine whether the following series converge. $$\sum_{k=1}^{\infty} \ln \left(\frac{k+2}{k+1}\right)$$

Short answer

Answer: The given series diverges.

Step-by-step solution

Identify the series and apply the Ratio Test

The given series can be written as: $$A_k = \ln\left(\frac{k+2}{k+1}\right)$$ To use the Ratio Test, we will consider the limit as k approaches infinity of the ratio between two consecutive terms: $$\lim_{k\to\infty} \left|\frac{A_{k+1}}{A_k}\right|$$

Calculate the ratio between two consecutive terms

We first find \(A_{k+1}\): $$A_{k+1} = \ln\left(\frac{(k+1)+2}{(k+1)+1}\right) = \ln\left(\frac{k+3}{k+2}\right)$$ Next, we calculate the ratio of \(A_{k+1}\) to \(A_k\): $$\left|\frac{A_{k+1}}{A_k}\right| = \left|\frac{\ln\left(\frac{k+3}{k+2}\right)}{\ln\left(\frac{k+2}{k+1}\right)}\right|$$

Calculate the limit of the ratio as k approaches infinity

We simplify the ratio and calculate the limit as k approaches infinity: $$\begin{aligned} \lim_{k\to\infty} \left|\frac{A_{k+1}}{A_k}\right| &= \lim_{k\to\infty} \left|\frac{\ln\left(\frac{k+3}{k+2}\right)}{\ln\left(\frac{k+2}{k+1}\right)}\right| \\ &= \lim_{k\to\infty} \left|\frac{\ln\frac{k+3}{k+2}- \ln\frac{k+2}{k+1}}{\ln\frac{k+2}{k+1}}\right| \\ &= \lim_{k\to\infty} \left|\frac{\ln\frac{(k+3)(k+1)}{(k+2)^2}}{\ln\frac{k+2}{k+1}}\right| \\ \end{aligned}$$

Analyze the limit of the ratio and determine the convergence

Now, we can observe the limit of the terms in the numerator and denominator separately as k approaches infinity and draw conclusions regarding the series: $$\begin{aligned} \lim_{k\to\infty} \ln\frac{(k+3)(k+1)}{(k+2)^2} &= \ln\frac{1\cdot 1}{1^2} = 0 \\ \lim_{k\to\infty} \ln\frac{k+2}{k+1} &= \ln\frac{\infty}{\infty} = \text{undefined} \\ \end{aligned}$$ Since the limit of the numerator approaches 0 while the limit of the denominator remains undefined, the overall limit of the ratio does not follow the rules of the Ratio Test. Therefore, the Ratio Test cannot be used to determine the convergence of this series. However, notice that the expression can be simplified: $$A_k = \ln\left(\frac{k+2}{k+1}\right)=\ln(k+2)-\ln(k+1)$$ So the series can be written as: $$\sum_{k=1}^{\infty} \left[\ln(k+2)-\ln(k+1)\right]$$ This is a telescoping series, meaning some terms get canceled out when we expand the series: $$\left[\ln(3)-\ln(2)\right] + \left[\ln(4)-\ln(3)\right] + \left[\ln(5)-\ln(4)\right] + \cdots$$ Upon canceling out terms over the series, we get: $$= \ln(3)-\ln(2) + \ln(4)-\ln(3) + \ln(5)-\ln(4) + \dots$$ As \(k\) approaches infinity, last term will be of the form \(\ln(k+2)-\ln(k+1)\) while \(\ln(k+1)\) will be remaining uncanceled. Since the logarithm function is unbounded as the argument approaches to infinity, the series is divergent.

Key concepts

Ratio Test

The Ratio Test is a common tool used to determine the convergence of infinite series. It involves taking the limit of the absolute value of the ratio of consecutive terms of the series. The test states that for a series \( \sum a_k \), if the limit \( \lim_ {k\to\infty} \left| \frac{a_{k+1}}{a_k} \right| = L \), then:

• If \( L < 1 \), the series converges absolutely.
• If \( L > 1 \) or \( L \) is infinite, the series diverges.
• If \( L = 1 \), the test is inconclusive.

In the original solution for the provided series, the Ratio Test was initially attempted. However, calculations show that the test yields an undefined result, suggesting the series might not be easily decided by this method alone. This prompts the need for alternative methods like telescoping series techniques.

Telescoping Series

A telescoping series is one where many terms cancel out, simplifying the series to a point where convergence can be more easily assessed. It usually occurs when the series terms can be expressed as a difference of successive terms in such a way that most intermediate terms cancel.

In our series, we noticed a form \( \ln(k+2) - \ln(k+1) \). When expanded, the series becomes an expression where most terms cancel out:

• \( \ln(3) - \ln(2) \)
• \( \ln(4) - \ln(3) \)
• \( \ln(5) - \ln(4) \)

The cancellation reduces the series substantially, leaving only the first and the last few terms, further aiding in assessing convergence. Ultimately, unbounded terms in telescoping series like this can signal divergence, as demonstrated in the solution.

Logarithmic Series

Logarithmic series refer to series expressions that include logarithmic functions. They oftentimes require specific consideration due to their properties. The series provided in the exercise, through its terms \( \ln\left(\frac{k+2}{k+1}\right) \), represents a sequence of logarithms.

Recognize that logarithmic behavior can inform convergence and divergence analyses. For large values of \( k \), logarithmic terms can provide slow-growing contributions, which makes logarithmic series intriguing when studying convergence. In our series, transformations like expressing terms as differences hinge on these behaviors. Recognizing this helps in potentially applying telescopic methods effectively.

Divergent Series

A divergent series is one that does not converge to a finite limit. In mathematical analysis, determining whether a series is divergent or convergent is crucial. A series might diverge when terms don’t tend towards zero quickly enough or if the cumulative sum increases indefinitely.

For example, in our exercise, after expressing the series as a telescoping series, it becomes clear that the cumulative logs don’t lead to a cancellation pointing to any finite sum. As such, ultimately, the behavior of the logarithms as \( k \to \infty \) shows the series' failure to find a finite bound, thus classifying it as divergent. Understanding what leads to divergence in different series types, including logarithmic and telescoping ones, is key to solving such problems.