Question

For each of the following sequences, if the divergence test applies, either state that \(\lim _{n \rightarrow \infty} a_{n}\) does not exist or find \(\lim _{n \rightarrow \infty} a_{n} .\) If the divergence test does not apply, state why. \(a_{n}=\frac{n}{\sqrt{3 n^{2}+2 n+1}}\)

Short answer

The limit is \( \frac{1}{\sqrt{3}} \) and the divergence test applies since the limit isn't zero.

Step-by-step solution

Identify the Sequence

The sequence given is \( a_{n} = \frac{n}{\sqrt{3n^2 + 2n + 1}} \). We are asked to check if \( \lim_{n \to \infty} a_n \) exists or if the divergence test can be used.

Simplify the Sequence

To find \( \lim_{n \to \infty} a_n \), simplify \( a_n \) by dividing the numerator and the denominator by \( n \): \[ a_{n} = \frac{n}{\sqrt{n^2(3 + \frac{2}{n} + \frac{1}{n^2})}} = \frac{1}{\sqrt{3 + \frac{2}{n} + \frac{1}{n^2}}} \]

Evaluate the Limit

As \( n \to \infty \), the terms \( \frac{2}{n} \) and \( \frac{1}{n^2} \) become negligible, so:\[ \lim_{n \to \infty} \frac{1}{\sqrt{3 + \frac{2}{n} + \frac{1}{n^2}}} = \frac{1}{\sqrt{3}} \]This limit evaluates to \( \frac{1}{\sqrt{3}} \).

Conclude on Divergence Test

The divergence test states if \( \lim_{n \to \infty} a_n eq 0 \), the series diverges. For this sequence, \( \lim_{n \to \infty} a_n = \frac{1}{\sqrt{3}} eq 0 \), so the series represented by these terms would diverge if we were considering a series rather than just the sequence.

Key concepts

Sequence Limit

When dealing with sequences in mathematics, an essential concept is finding their limit. The sequence limit, denoted by \( \lim_{n \to \infty} a_n \), gives us an idea of what value the sequence approaches as \( n \) becomes infinitely large. In the sequence \( a_n = \frac{n}{\sqrt{3n^2 + 2n + 1}} \), our goal is to determine whether a limit exists as \( n \) tends toward infinity.
To compute this, we focus on simplifying the expression which often involves manipulating terms to reveal dominant behaviors. Here, simplifying the sequence lets us identify if and when these terms settle towards a specific number. The result tells us whether the sequence converges to this limit or not. If a limit exists and is finite, the sequence converges; otherwise, it may diverge or not have a simple result.
Knowing how to find sequence limits is crucial, especially in calculus, as it lays the foundation for deeper understandings of function behaviors, continuity, and Series Divergence.

Simplifying Sequences

Simplifying sequences is a key strategy to make it easier to find the limit or to make other analysis simpler. By reducing a complex sequence to a simpler form, it becomes more straightforward to observe the sequence's behavior.
In our example with \( a_n = \frac{n}{\sqrt{3n^2 + 2n + 1}} \), simplification played a central role. We started by dividing both the numerator and the denominator by \( n \), resulting in \( \frac{1}{\sqrt{3 + \frac{2}{n} + \frac{1}{n^2}}} \). This step helps to highlight the primary components that dictate how the sequence behaves as \( n \to \infty \).
By doing such simplifications, you're able to focus on dominant terms which directly influence the sequence's limit. In this case, irrelevant terms like \( \frac{2}{n} \) and \( \frac{1}{n^2} \) become negligible, clarifying the true limit is \( \frac{1}{\sqrt{3}} \). Such simplification is a powerful tool not only for understanding an individual sequence but also for larger mathematical explorations and series analyses.

Series Divergence

Understanding whether a series diverges is a fundamental concept in calculus and the study of series. It's essential for math learners to recognize the divergence test, which helps in understanding the behavior of a series when regarded collectively over infinite terms.
The Divergence Test suggests that if \( \lim_{n \to \infty} a_n eq 0 \), then the series \( \sum_{n=1}^{\infty} a_n \) diverges. This means the series doesn’t settle at a particular value but instead grows infinitely larger or behaves erratically, failing to converge.
For the sequence \( a_n = \frac{n}{\sqrt{3n^2 + 2n + 1}} \), the limit resolved to \( \frac{1}{\sqrt{3}} \). Since this limit is not zero, a series formed from this sequence would diverge per the Divergence Test. Realizing series divergence is vital, as it informs mathematicians and students that such a series doesn’t provide a summed value and is important in various applications like stability analysis in differential equations or determining the behavior in statistical series.