Problem 43
Question
Determine the convergence or divergence of the sequence with the given \(n\) th term. If the sequence converges, find its limit. \(a_{n}=\frac{3^{n}}{4^{n}}\)
Step-by-Step Solution
Verified Answer
The given sequence converges and as \(n\) approaches infinity, the sequence tends to 0.
1Step 1: Simplify the Sequence
Start by simplifying the sequence. Note that \(a_{n}=\frac{3^{n}}{4^{n}}\) can be rewritten as \(a_{n}=(\frac{3}{4})^{n}\), this makes it easier to visualize and to compute the next steps.
2Step 2: Apply the Ratio Test
Apply the Ratio test by taking the limit as \(n\) goes to infinity of the ratio of successive terms \(\frac{a_{n+1}}{a_{n}}\). In this case this becomes \(lim_{n \to +\infty}\frac{ (\frac{3}{4})^{n+1}}{(\frac{3}{4})^{n}}\). Simplify this ratio to \(lim_{n \to +\infty}\frac{3}{4}\), being the ratio of the constants, independent of \(n\), it doesn't tend to infinity.
3Step 3: Compare with 1
According to the ratio test, if the limit of the ratio is less than 1, which is clearly the case here as \(\frac{3}{4} < 1\), then the series converge. This implies that the given sequence also converges.
4Step 4: Find the limit of the sequence
As \(n\) tends to infinity, \((\frac{3}{4})^{n}\) will tend to 0 as \(3/4<1\), thus it ensures that raising it to higher powers will make the term increasingly smaller. Hence, the limit of the sequence as \(n\) approaches infinity is 0.
Key Concepts
Limit of a SequenceRatio TestInfinite Series
Limit of a Sequence
Understanding the limit of a sequence is crucial when analyzing whether a sequence converges to a particular value or diverges. A sequence is essentially a list of numbers generated by some mathematical rule. In formal terms, the limit of a sequence \(a_n\) as \(n\) approaches infinity is the value that \(a_n\) approaches as \(n\) becomes very large. If such a limit exists and is finite, we say that the sequence converges; if the limit does not exist or is infinite, the sequence diverges.
For example, the sequence given by \(a_n = \frac{3^n}{4^n}\) simplifies to \(a_n = \left(\frac{3}{4}\right)^n\). As \(n\) grows, the value of \(a_n\) gets smaller because \(\frac{3}{4}\) is less than 1. Intuitively, you're repeatedly multiplying by a fraction that's less than one, which diminishes the product over time. Mathematically, we can express this as \(\lim_{n \to \infty} \left(\frac{3}{4}\right)^n = 0\), indicating that the sequence converges to 0 as \(n\) tends to infinity.
For example, the sequence given by \(a_n = \frac{3^n}{4^n}\) simplifies to \(a_n = \left(\frac{3}{4}\right)^n\). As \(n\) grows, the value of \(a_n\) gets smaller because \(\frac{3}{4}\) is less than 1. Intuitively, you're repeatedly multiplying by a fraction that's less than one, which diminishes the product over time. Mathematically, we can express this as \(\lim_{n \to \infty} \left(\frac{3}{4}\right)^n = 0\), indicating that the sequence converges to 0 as \(n\) tends to infinity.
Ratio Test
The Ratio Test is a valuable tool for determining the convergence or divergence of infinite series and by extension can also be applied to sequences. The test looks at the limit of the ratio of successive terms in the series (or sequence), \(\lim_{n \to \infty} \frac{a_{n+1}}{a_{n}}\).
To apply the ratio test for our sequence \(a_{n} = \left(\frac{3}{4}\right)^n\), we examine \(\lim_{n \to \infty} \frac{\left(\frac{3}{4}\right)^{n+1}}{\left(\frac{3}{4}\right)^n}\) which simplifies to \(\lim_{n \to \infty} \frac{3}{4}\). Since this ratio is constant and less than 1, the test tells us that the sequence converges. Note that if the limit of the ratio exceeds 1, or oscillates without approaching a finite limit, the sequence or series may diverge.
To apply the ratio test for our sequence \(a_{n} = \left(\frac{3}{4}\right)^n\), we examine \(\lim_{n \to \infty} \frac{\left(\frac{3}{4}\right)^{n+1}}{\left(\frac{3}{4}\right)^n}\) which simplifies to \(\lim_{n \to \infty} \frac{3}{4}\). Since this ratio is constant and less than 1, the test tells us that the sequence converges. Note that if the limit of the ratio exceeds 1, or oscillates without approaching a finite limit, the sequence or series may diverge.
Infinite Series
An infinite series is the sum of the terms of an infinite sequence. In simpler terms, it's what you get when you try to add up an endless list of numbers. The convergence or divergence of an infinite series is intimately connected to the behavior of the sequence it's derived from.
It's valuable to recognize that a sequence can converge even when its corresponding series does not. For instance, even though \(a_n = \left(\frac{3}{4}\right)^n\) converges to 0, it doesn't necessarily mean that the series \(\sum_{n=0}^{\infty} \left(\frac{3}{4}\right)^n\) converges. However, in this particular example, because each term gets smaller and smaller and the ratio between successive terms is consistently less than 1, this particular series does indeed converge (specifically, it is a geometric series with a common ratio less than 1). Recognizing the type of series, such as geometric, can be exceptionally helpful in determining the overall behavior of the sum.
It's valuable to recognize that a sequence can converge even when its corresponding series does not. For instance, even though \(a_n = \left(\frac{3}{4}\right)^n\) converges to 0, it doesn't necessarily mean that the series \(\sum_{n=0}^{\infty} \left(\frac{3}{4}\right)^n\) converges. However, in this particular example, because each term gets smaller and smaller and the ratio between successive terms is consistently less than 1, this particular series does indeed converge (specifically, it is a geometric series with a common ratio less than 1). Recognizing the type of series, such as geometric, can be exceptionally helpful in determining the overall behavior of the sum.
Other exercises in this chapter
Problem 42
Determine whether the series converges conditionally or absolutely, or diverges. $$ \sum_{n=1}^{\infty} \frac{\sin [(2 n-1) \pi / 2]}{n} $$
View solution Problem 43
Find the sum of the convergent series. $$ \sum_{n=0}^{\infty}\left(\frac{1}{2^{n}}-\frac{1}{3^{n}}\right) $$
View solution Problem 43
Define a power series centered at \(c\).
View solution Problem 43
Use the Direct Comparison Test to determine the convergence or divergence of the series. $$ \sum_{n=0}^{\infty} \frac{1}{3^{n}+1} $$
View solution