Question

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

Short answer

The series is a \( p \)-series with \( p = 0.386 \) and diverges (\( p < 1 \)).

Step-by-step solution

Identify the structure of the sum

Observing the given series \( \sum_{n=1}^{\infty} n 2^{-2 \ln n} \), we can recognize that the term \( 2^{-2 \ln n} \) can be rewritten to match a form similar to a power of \( n \). We aim to express each term in the form \( n^{-p} \).

Simplify the exponent

Notice that \( 2^{-2 \ln n} \) can be rewritten by using properties of logarithms and exponents. Recall that \( a^{b} = e^{b \ln a} \), so \( 2^{-2 \ln n} = \left(e^{\ln 2}\right)^{-2 \ln n} = e^{-2 \ln(n) \ln(2)} \). Therefore, the term becomes \( e^{-\ln n^{2\ln 2}} = n^{-2 \ln 2} \).

Reformulate the series

Using the simplification from Step 2, the given series now takes the form \( \sum_{n=1}^{\infty} n \cdot n^{-2 \ln 2} = \sum_{n=1}^{\infty} n^{1 - 2 \ln 2} \).

Determine the convergence

Recognize that the series \( \sum_{n=1}^{\infty} n^{1 - 2 \ln 2} \) is a \( p \)-series with \( p = 2 \ln 2 - 1 \). A \( p \)-series \( \sum_{n=1}^{\infty} n^{-p} \) converges if \( p > 1 \). Calculate \( 2 \ln 2 \) which is approximately \( 1.386 \), so \( 2 \ln 2 - 1 = 0.386 \). Since \( 0.386 < 1 \), the series diverges.

Key concepts

Convergence

Convergence is about determining if a series sums to a specific number or not. In math, particularly with infinite series, convergence means that as you add more and more terms, the sum approaches a finite value. This is a key topic in calculus and analysis.
To figure out if a series converges, we often use tests such as the p-series test, which tells us if a series of the form \( \sum_{n=1}^{\infty} n^{-p} \) will converge. The rule is simple: if \( p > 1 \), the series converges, and if \( p \leq 1 \), it diverges.
Understanding convergence helps in many areas of math and science because it can predict the behavior of sequences. It tells us if the sum of infinite terms results in a finite answer or not.

Properties of Logarithms

Logarithms help us simplify complex mathematical expressions. The properties of logarithms allow us to manipulate exponentials and they work hand-in-hand with exponents.
Here are some useful properties:

• Product Rule: \( \log_b(xy) = \log_b(x) + \log_b(y) \)
• Quotient Rule: \( \log_b\left(\frac{x}{y}\right) = \log_b(x) - \log_b(y) \)
• Power Rule: \( \log_b(x^a) = a\log_b(x) \)
• Change of Base Formula: \( \log_b(x) = \frac{\log_k(x)}{\log_k(b)} \) for any positive base \( k \)

In our exercise, these properties are used to rewrite \( 2^{-2 \ln n} = n^{-2 \ln 2} \). This transformation is crucial for analyzing the convergence of the series as a p-series.

Exponents

Exponents are about multiplying a number by itself a certain number of times. They are shown as a small number above and to the right of a base number, like \( n^p \).
In our case, understanding exponents involves transforming something like \( 2^{-2 \ln n} \) to a simple form that can connect to a p-series. One key idea is the formula \( a^{b} = e^{b \ln a} \), which converts powers and bases to a form involving logarithms. This helps in the simplification and understanding of the series' terms.
A reminder:

• Negative Exponent: \( x^{-a} = \frac{1}{x^a} \)
• Any number to the power of zero is 1: \( x^0 = 1 \)
• Multiplying similar bases adds exponents: \( x^a \, \times \, x^b = x^{a+b} \)

Exponents are versatile and knowing their properties makes complicated expressions easier to handle.

Series Decomposition

Series decomposition involves breaking down a complex series into simpler parts or steps. It's like disassembling a puzzle to see how each piece fits.
For the exercise, the series is initially given as \( \sum_{n=1}^{\infty} n \, 2^{-2 \ln n} \). Notice how it’s not in the classic p-series format. So, we decompose it into something we understand better. This simplification involves converting the exponent and separating the series into a sum of powers to check convergence. Let's break it down:

• Transform the series so that each term is expressed as \( n^{-p} \) using logarithmic and exponential properties.
• Identify the resulting power \( p \) and determine if \( p > 1 \) for convergence.

This process makes handling and solving infinite series much more manageable, allowing us to apply known convergence tests.