Problem 45 - Thomas Calculus

Question

Which of the sequences \(\left\\{a_{n}\right\\}\) converge, and which diverge? Find the limit of each convergent sequence. $$ a_{n}=\sqrt{\frac{2 n}{n+1}} $$

Step-by-Step Solution

Answer

The sequence converges to \(\sqrt{2}\).

Step 1: Identify the Sequence Form

We are given the sequence \(a_n = \sqrt{\frac{2n}{n+1}}\). The task is to determine if this sequence converges or diverges, and if it converges, to find its limit.

Step 2: Simplify the Expression

To analyze the behavior as \(n\) approaches infinity, simplify \(a_n = \sqrt{\frac{2n}{n+1}}\) by dividing both the numerator and the denominator inside the square root by \(n\). This gives \(a_n = \sqrt{\frac{2}{1 + \frac{1}{n}}}\).

Step 3: Analyze the Limit

As \(n\) approaches infinity, \(\frac{1}{n}\) tends to 0. Therefore, the expression \(1 + \frac{1}{n}\) approaches 1, and \(\frac{2}{1 + \frac{1}{n}}\) approaches 2. So, \(a_n = \sqrt{\frac{2}{1 + \frac{1}{n}}}\) approaches \(\sqrt{2}\).

Step 4: Conclude Convergence

Since as \(n\) approaches infinity, the sequence \(a_n = \sqrt{\frac{2}{1 + \frac{1}{n}}}\) tends towards \(\sqrt{2}\), the sequence converges. The limit of the sequence is \(\sqrt{2}\).

Key Concepts

Limit of a Sequence

When we talk about the limit of a sequence, we are describing what the sequence approaches as the term number grows infinitely large. Essentially, a sequence has a limit when all its terms become arbitrarily close to a specific number. In the given exercise, the sequence is \(a_n = \sqrt{\frac{2n}{n+1}}\). As \(n\) increases, we see that the terms in the sequence approach \(\sqrt{2}\), indicating convergence.

• The limit of a sequence \(\{a_n\}\) as \(n\) goes to infinity can be mathematically expressed as \(\lim_{{n \to \infty}} a_n = L\), where \(L\) is a real number.
• If a sequence has a limit, it converges to that limit.
• For our sequence, we simplified and found \(\lim_{{n \to \infty}} \sqrt{\frac{2}{1 + \frac{1}{n}}} = \sqrt{2}\).

Divergence

Not all sequences converge. When a sequence does not approach any particular value as its terms grow larger, it is said to diverge. Divergence means that the terms of the sequence keep moving away from a particular value or oscillate indefinitely without settling to a limit.

• A simple example is the sequence \(b_n = n\), which becomes infinitely large as \(n\) increases; thus, it diverges.
• Another example is a sequence that jumps between values, like \((-1)^n\), which alternates between -1 and 1 and also diverges.
• In the exercise provided, \(a_n\) converges to \(\sqrt{2}\), so it does not exhibit divergence.

Infinite Limits

When talking about infinite limits, we refer to sequences whose terms grow larger and larger without bound. Such sequences do not settle on a finite number and instead head toward infinity.

• We express an infinite limit scenario with notation like \(\lim_{{n \to \infty}} a_n = \infty\).
• Infinite limits often suggest divergence because the sequence does not approach any real number.
• For instance, in the exercise, we calculated limit values that confirmed convergence to \(\sqrt{2}\), rather than implying infinite behavior.

Understanding infinite limits helps differentiate between continuous growth patterns and those that stabilize, which is vital in mathematical analyses of sequences.