Question

Express the following sums as \(p\) -series and determine whether each converges. \(\sum_{n=1}^{\infty} n 3^{-2 \ln n}\)

Short answer

The series converges as it is a p-series with \(p = 1.1972 > 1\).

Step-by-step solution

Simplify the Term

First, observe the term in the series: \(n 3^{-2 \ln n}\). We can rewrite it with a common exponential identity. Recall that \(3^{-2 \ln n} = e^{-2 \ln n \ln 3}\). Simplifying further, we get \(n e^{-2 \ln n \ln 3} = n n^{-2 \ln 3} = n^{1 - 2 \ln 3}\). Thus, the general form for the term is \(n^{1 - 2 \ln 3}\).

Identify the p-Series

A p-series can be expressed in the form \(\sum_{n=1}^{\infty} \frac{1}{n^p}\). Comparing it with our term, we need to express the form as \(\sum_{n=1}^{\infty} n^{-p}\), where \(p = 2 \ln 3 - 1\). Thus, our series is \(\sum_{n=1}^{\infty} n^{-(2 \ln 3 - 1)}\).

Determine Convergence Criteria

For a p-series \(\sum_{n=1}^{\infty} n^{-p}\) to converge, the condition is that \(p > 1\). We need to check if \(2 \ln 3 - 1 > 1\).

Evaluate Convergence Condition

Calculate \(\ln 3\) using an approximation or calculator, \(\ln 3 \approx 1.0986\). So, \(2 \ln 3 \approx 2 \times 1.0986 = 2.1972\). Therefore, \(2 \ln 3 - 1 = 1.1972\). Since \(1.1972 > 1\), the series converges.

Key concepts

Convergence of Series

In mathematics, the convergence of a series refers to whether the sum of its terms reaches a finite value as more terms are added. When this happens, we say the series converges. For a series to converge, especially a p-series, mathematical criteria must be fulfilled. A p-series takes the form \( \sum_{n=1}^{\infty} \frac{1}{n^p} \), and its convergence relies on the value of the exponent \( p \).

• If \( p > 1 \), the series converges, meaning it adds up to a finite number.
• If \( p \leq 1 \), the series diverges, which means it doesn't settle into a finite sum, essentially becoming infinite.

Checking for series convergence involves comparing it to these rules. This provides a simplified method to determine if a series will behave predictably as you sum infinitely many terms.

Exponential Functions

Exponential functions are functions of the form \( f(x) = a^x \), where \( a \) is a positive real number. These functions show up in many areas of mathematics due to their unique properties and behavior with respect to growth and decay. A key point of exponential functions is how they handle multiplicative changes.
One special case is the natural exponential function \( e^x \), where \( e \) is an irrational constant approximately equal to 2.718. This function grows at a rate proportional to its current value, making it central to calculus and mathematical modeling.
In our original exercise, we simplified an exponential expression using the identity \( a^{b \ln c} = c^{\ln a^b} \). This converted our given series expression into a power of \( n \), paving the way to express it as a p-series. The simplification process is crucial to applying convergence criteria accurately.

Logarithmic Functions

Logarithmic functions are inverses of exponential functions. When you have an exponential function \( a^x = y \), the equivalent logarithmic function is \( \log_a (y) = x \). Specifically, the natural logarithm, denoted \( \ln \), uses the constant \( e \) as its base and is widely used due to its mathematical characteristics.
A well-known property of logarithms relates to their ability to transform multiplication into addition, thanks to the identity \( \log_b(xy) = \log_b(x) + \log_b(y) \). This property is integral in simplifying complex equations, especially when exponentials are involved, as seen in the exercise. We used logarithmic properties to break down and simplify the term \( n 3^{-2 \ln n} \) into a more manageable expression for determining convergence.
Understanding logarithmic functions allows us to work more efficiently with exponential forms, and they emerge naturally in formulas across different scientific disciplines.