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The comparison test is a handy tool when analyzing whether an infinite series converges or diverges. It works by comparing a complex series to a simpler, well-understood series. If the simpler series behaves a certain way (either converging or diverging), the series you are testing likely behaves the same way.
In this exercise, we use the comparison test to analyze the series \( \sum_{n=1}^{\infty} \frac{1}{n+3^{n}} \). By comparing it to the series \( \sum_{n=1}^{\infty} \frac{1}{3^n} \), which we know converges, we can conclude the same for our series since \( \frac{1}{n+3^n} \leq \frac{1}{3^n} \) for all \( n \geq 1 \).
To use the comparison test effectively:

• First, identify a comparison series that is similar in structure.
• Then, ensure the terms of your complicated series are always less than or equal to the terms of this simpler convergent series.

Geometric series are a class of infinite series where each subsequent term is obtained by multiplying the previous term by a constant ratio \( r \). In mathematical form, a geometric series looks like: \[ a + ar + ar^2 + ar^3 + \cdots \]where \( a \) is the initial term, and \( r \) is the common ratio between terms.
A geometric series converges if the absolute value of the ratio \( |r| \) is less than 1. Otherwise, it diverges.
For example, in this exercise, we compared our infinite series to the geometric series \( \sum_{n=1}^{\infty} \frac{1}{3^n} \) with \( a = \frac{1}{3} \) and \( r = \frac{1}{3} \). Because the ratio \( |\frac{1}{3}| \) is less than one, this series converges. This provided a basis for our comparison test.

An infinite series is essentially the sum of an infinite sequence of terms. To determine whether an infinite series like \( \sum_{n=1}^{\infty} \frac{1}{n+3^{n}} \) converges or diverges, we need to establish whether the sum reaches a definite, finite number as more terms are added, i.e., as \( n \to \infty \).
An infinite series converges if the sequence of partial sums approaches a single finite limit. Otherwise, it diverges, potentially growing infinitely large or oscillating without settling.
In this exercise, we see an infinite series where the denominator contains an exponential term \( 3^n \), which grows quickly, aiding in the evaluation of convergence using tests like the comparison test. Understanding the nature of the denominator can significantly inform how we approach solving for convergence.

• Convergence: Implies the sum of infinite terms remains finite.
• Divergence: Indicates that the sum grows indefinitely or lacks a fixed limit.