Question

Determine whether the sequence converges or diverges. If convergent, give the limit of the sequence. $$\left\\{a_{n}\right\\}=\left\\{\frac{1.1^{n}}{n}\right\\}$$

Short answer

The sequence diverges.

Step-by-step solution

Understand the Problem

We are given a sequence \( \{a_n\} = \left\{\frac{1.1^n}{n}\right\} \). The task is to determine whether this sequence converges (approaches a specific value as \( n \) increases indefinitely) or diverges (does not approach a specific value).

Check the Form of the Sequence

The given sequence is in the form \( a_n = \frac{1.1^n}{n} \), where the numerator \( 1.1^n \) represents an exponential term and the denominator \( n \) represents a linear term.

Analyze the Growth of Terms

As \( n \) becomes very large, \( 1.1^n \) grows exponentially because its base is greater than 1. On the other hand, \( n \) grows linearly. Therefore, the numerator increases much more rapidly than the denominator.

Determine Convergence or Divergence

Since the exponential term in the numerator grows faster than the linear term in the denominator as \( n \to \infty \), the sequence \( \{a_n\} = \left\{\frac{1.1^n}{n}\right\} \) will also grow indefinitely. Hence, the sequence diverges.

Key concepts

Exponential Growth

Exponential growth occurs when a quantity increases by a constant factor over equal intervals of time. This is often seen in processes like population growth, interest accumulation, and, as in our case, sequences with exponential factors.

• For our sequence \(a_n = \frac{1.1^n}{n}\), the term \(1.1^n\) reflects exponential growth.
• This is because it increases rapidly as \(n\) gets larger, due to the base 1.1 being greater than 1.

In essence, exponential growth in a sequence causes each term to become significantly larger than the previous term as \(n\) continues to rise. And when comparing this growth against linear or slower-growing terms, it usually dominates due to its rapid escalation over time.

Linear Growth

Linear growth describes a situation where a quantity increases by a constant amount over equal intervals of time. This type of growth is simpler and more predictable than exponential growth.

• In the sequence \(a_n = \frac{1.1^n}{n}\), the denominator \(n\) shows linear growth.
• This means \(n\) increases steadily and predictably as each term progresses.

However, when linear growth is pitted against exponential growth, as in our sequence, it will often be overpowered. The linear denominator cannot counteract the rapid increase of the exponential numerator, which is crucial when determining if the sequence will converge or diverge.

Limits

In the context of sequences, a limit is a value that the terms of a sequence "approach" as the index \(n\) goes to infinity. Not every sequence has a limit. When it does, the sequence is said to converge.

• If a sequence converges, it will settle at a particular number or state, regardless of whether the sequence was initially increasing or decreasing.
• To find a limit, we analyze the behavior of the sequence as \(n\) becomes very large.

For the given sequence \(a_n = \frac{1.1^n}{n}\), the exponential growth of the numerator prevents the terms from approaching a specific value. Therefore, this sequence does not converge, so it does not possess a limit.

Divergence

Divergence in sequences refers to a situation where the terms of the sequence do not settle into a limit. Instead, they continue to increase, decrease, or oscillate without approaching a fixed value as \(n\) goes to infinity.

• Divergent sequences grow without bound, or they might oscillate endlessly.
• In the case of our sequence \(a_n = \frac{1.1^n}{n}\), the rapid growth of the numerator leads the sequence to increase indefinitely.

The exponential numerator vastly outpaces the linear growth in the denominator, ensuring there is no finite limit the sequence tends toward. Hence, the sequence diverges, illustrating how exponential terms can dominate the overall behavior of a sequence.