Problem 79 - Essential Calculus

Question

Determine the convergence or divergence of the series. $$ \frac{1}{201}+\frac{1}{204}+\frac{1}{209}+\frac{1}{216}+\cdots \cdot $$

Step-by-Step Solution

Answer

The given series converges.

Step 1: Identify the Series

Firstly, recognise the pattern of the input series, where the denominator for each term seems to follow the rule \( n^3 + 2n \). Thus, the series can be re-written as \( \sum _{n=3} ^{\infty} \frac{1}{n^3+2n} \)

Step 2: Select Appropriate Test

To assess the convergence or divergence of the sum of the series, it is suggested to use the comparison test. The comparison test would be favorable since we could easily compare it against a known convergent series - \( \frac{1}{n^3} \) which is a p-series with \( p > 1 \), hence convergent.

Step 3: Apply the Comparison Test

To apply the comparison test, we compare our given series with the known convergent series. Since \( \frac{1}{n^3+2n} \leq \frac{1}{n^3} \) for all \( n \geq 3 \), and we know that the series \( \sum \frac{1}{n^3} \) is a convergent p-series, we can conclude via the comparison test that our original series \( \sum _{n=3} ^{\infty} \frac{1}{n^3+2n} \) converges.

Key Concepts

Comparison Test

The Comparison Test is a handy tool when determining the convergence or divergence of an infinite series. It involves comparing a series whose convergence we know to a series whose convergence we're trying to determine. Here's how it works: you select a simpler known series that closely resembles your given series. If you can show that each term of your series is smaller than the corresponding term of a known converging series, and this known series converges, then your series also converges.

• For example, with the series in question, each term in the form of \( \frac{1}{n^3 + 2n} \) was compared to \( \frac{1}{n^3} \) from a known convergent p-series.
• Think of it as having a buddy next to you in a race. If you know your buddy will reach the finish line and you never get behind them, you'll reach the goal too.

Remember, finding a known convergent series that resembles your series is paramount. The Comparison Test simplifies the process by leveraging our existing knowledge of other series.

Series Convergence

Series Convergence is a central concept in calculus where we determine whether the sum of an infinite series of numbers approaches a fixed number. It's like trying to figure out if the sum of an endless list of numbers will eventually settle on one total. If it does, we say the series converges; if not, it diverges.

• To explore convergence, several tests can be applied, such as the Comparison Test, Ratio Test, and the Integral Test.
• Convergence is determined using tests as it ultimately tells us whether adding infinite terms of the series will yield a finite result.

In this particular problem, the Comparison Test was applied to assess convergence. Since the terms diminished quickly enough, the infinite sum stayed finite, depicting convergence. It's this shrinking behavior of terms that often indicates a series converges.

P-Series

A p-series is a particular kind of series. Its general form is \( \sum \frac{1}{n^p} \), where \( n \) starts at 1 (or sometimes another number) and continues indefinitely. The behavior of a p-series depends on the value of \( p \). This is a straightforward, yet powerful form of series that provides a base for many convergence tests:

• If \( p > 1 \), the p-series converges.
• If \( p \leq 1 \), the p-series diverges.

In our context, a p-series with \( p = 3 \) served as a reference in the Comparison Test. Knowing its properties allowed us to efficiently determine the convergence of the more complex series \( \sum \frac{1}{n^3 + 2n} \). Understanding p-series is essential as they form the foundation for many concepts in series convergence.