Question

Evaluate the series or state that it diverges. $$\sum_{k=2}^{\infty} \frac{\ln \left((k+1) k^{-1}\right)}{(\ln k) \ln (k+1)}$$

Short answer

Answer: The given series converges.

Step-by-step solution

Simplify the series term

Let's first simplify the term inside the series: $$\frac{\ln \left((k+1) k^{-1}\right)}{(\ln k) \ln (k+1)} = \frac{\ln \left(\frac{k+1}{k}\right)}{(\ln k) \ln (k+1)}$$ Now, we can use the property of logarithms: \(\ln(a/b) = \ln(a) - \ln(b)\). So we get: $$\frac{\ln \left(\frac{k+1}{k}\right)}{(\ln k) \ln (k+1)} = \frac{\ln(k+1) - \ln(k)}{(\ln k) \ln(k+1)}$$

Compare the series to a known convergent series

Now, we will compare this series to the harmonic series \(\sum_{k=2}^{\infty} \frac{1}{k}\). We observe that if \(\frac{\ln(k+1) - \ln(k)}{(\ln k) \ln(k+1)} \leq \frac{1}{k}\), then our series converges, because the harmonic series converges. Now, we can manipulate the inequality: $$\frac{\ln(k+1) - \ln(k)}{(\ln k) \ln(k+1)} \leq \frac{1}{k}$$ $$\frac{\ln(k+1) - \ln(k)}{\ln k} \leq \frac{\ln(k+1)}{k}$$ $$k \leq k + 1 - \frac{k}{\ln k}$$ Simplify the equation: $$0 \leq 1 - \frac{k}{\ln k}$$ For all values of \(k \geq 2\), the inequality is true, because \(\ln k\) is always positive. Therefore, the given series converges.

Determine the convergence of the given series

Since we know that the series is convergent, the next step is to find its value. However, it is important to mention that our comparison test only established convergence and not the exact value of the convergent series. Therefore, although we can conclude that the given series converges, we cannot determine its exact value using this method. In conclusion, the given series $$\sum_{k=2}^{\infty} \frac{\ln \left(\frac{k+1}{k}\right)}{(\ln k) \ln(k+1)}$$ converges, but we cannot determine its exact value using the comparison test.

Key concepts

Comparison Test

The Comparison Test is a useful way to determine whether a series converges or diverges. It involves comparing the series in question to another series whose behavior (convergence or divergence) is already known.
When using the comparison test, identify a series whose terms are either less than or equal to those of a known convergent series or greater than or equal to those of a known divergent series for all terms beyond a certain point.

• If the terms of your series are less than or equal to the terms of a known convergent series, then your series also converges.
• If the terms of your series are greater than or equal to the terms of a known divergent series, then your series also diverges.

In the given exercise, we used the harmonic series, which is a known divergent series for comparison. By establishing a relationship showing that the series in question behaves similarly to the harmonic series, we deduce its convergence characteristics.

Logarithms

Logarithms are mathematical expressions that help simplify complicated multiplication and division tasks into addition and subtraction tasks. The logarithm to the base 10 of a number is a power to which the base 10 must be raised to produce that number.
In mathematical notation, if \(a^x = b\), then \(\log_a(b) = x\).
In the exercise, we simplify the expression \(\ln\left(\frac{k+1}{k}\right)\) by applying the property: \(\ln(a/b) = \ln(a) - \ln(b)\). This transformation makes it easier to apply comparison tests to analyze series convergence. Logarithmic properties like this one are vital tools in calculus for transforming expressions and unraveling the complex behavior of functions.

Harmonic Series

The Harmonic Series is a crucial concept in mathematical analysis, consisting of the sum of reciprocals of all natural numbers: \[ \sum_{k=1}^{\infty} \frac{1}{k} \]Despite its simple form, this series diverges, meaning that it grows without bound.In the context of series convergence tests, the harmonic series serves as a benchmark to which other series can be compared. If a series behaves like the harmonic series, it means it doesn't settle to a finite value, leading to divergence. However, if a series has terms consistently smaller than those of the harmonic series, it suggests that the series might converge. In our exercise, we leveraged the comparison with the harmonic series to conclude the convergence of the given series.

Calculus

Calculus is the branch of mathematics that studies continuous change, much like geometry studies shape or algebra studies operations. The fundamental components of calculus include derivatives, integrals, limits, and infinite series, all of which are essential tools in understanding dynamic systems. Infinite series, like the one in the problem, are a key area in calculus. They help represent functions as sums of infinitely many terms, offering significant insights in fields such as physics and engineering. Calculus uses tools like the comparison test to determine whether these series sum up to a finite value, which is crucial in ensuring the stability of solutions in practical applications. Understanding how calculus can describe series like the one given in the exercise equips students to tackle broader mathematical and real-world challenges.