Problem 68 - Single Variable Calculus

Question

Find the limit of the sequence $$ \\{\sqrt{2}, \sqrt{2 \sqrt{2}}, \sqrt{2 \sqrt{2 \sqrt{2}}}, \ldots\\} $$ Hint: Show that \(a_{n}=2^{\left(2^{n}-1\right) / 2^{n}}=2^{1-1 / 2^{n}}\).

Step-by-Step Solution

Answer

In summary, the sequence given is $$ \\{\sqrt{2}, \sqrt{2 \sqrt{2}}, \sqrt{2 \sqrt{2 \sqrt{2}}}, \ldots\\} $$ and its general term is \(a_n = 2^{1 - \frac{1}{2^{n-1}}}\). The limit of the sequence as n approaches infinity is 2.

Step 1: Identify the pattern and find \(a_n\)

Notice that $$ a_1 = \sqrt{2} = 2^{1/2}, $$ $$ a_2 = \sqrt{2 \sqrt{2}} = \sqrt{2}^2 = 2^{(1 + 1)/2}, $$ $$ a_3 = \sqrt{2 \sqrt{2 \sqrt{2}}} = \sqrt{2}^4 = 2^{(2^2)/2}, $$ and so on. In general, we can see that the exponent of 2 in the nth term \(a_n\) is obtained by taking the sum of the first n powers of 1/2: $$ a_n = 2^{\frac{1}{2} + \frac{1}{2^2} + \cdots + \frac{1}{2^n}}. $$

Step 2: Express the sum as a geometric series

Now, we rewrite the sum of the exponents using the geometric series formula. The sum of the first n powers of 1/2 is given by: $$ S_n = \frac{1 - \frac{1}{2^n}}{1 - \frac{1}{2}} = 2 - \frac{1}{2^{n-1}}. $$

Step 3: Use the hint to express \(a_n\) in terms of \(2^n\)

We can now express \(a_n\) in terms of \(2^n\) using the hint: $$ a_n = 2^{S_n} = 2^{\left(2 - \frac{1}{2^{n-1}}\right)} = 2^{1 - \frac{1}{2^{n-1}}}. $$ Thus, we found the general form for \(a_n\) as the hint suggested: $$ a_n = 2^{1 - \frac{1}{2^{n-1}}}. $$

Step 4: Find the limit

Now that we have a general expression for the nth term, we can find the limit as n approaches infinity: $$ \lim_{n \to \infty} a_n = \lim_{n \to \infty} 2^{1 - \frac{1}{2^{n-1}}}. $$ As n approaches infinity, we see that the term in the exponent tends towards zero: $$ \lim_{n \to \infty} 1 - \frac{1}{2^{n-1}} = 1. $$ Hence, the limit of the nth term is: $$ \lim_{n \to \infty} a_n = 2^1 = 2. $$ The limit of the sequence is 2.

Key Concepts

Convergence of Sequences

Understanding the concept of convergence in sequences is fundamental when you're dealing with limits. A sequence converges when its terms get closer and closer to a certain value as the sequence progresses indefinitely. This certain value is known as the limit of the sequence.

For the given sequence where each term is constructed using nested square roots of 2, determining convergence is about finding a single value that the terms approach as the sequence goes on to infinity. In the case of our sequence, as the number of nested square roots increases, the terms of the sequence approach closer to the number 2. This behavior of approaching a limit can sometimes be counterintuitive, especially with sequences featuring complex patterns or operations, just like nested roots.

Another takeaway is that not all sequences converge. Some may diverge, meaning they don't settle down to any single value, or may oscillate between different values. Recognizing whether a sequence converges and finding its limit involves understanding its underlying pattern, using certain properties of limits, or applying theorems from calculus. For our exercise, we see that the limit exists and is found to be the number 2.

Geometric Series

Unlocking Patterns with Geometric Series

When we encounter a sequence with terms that have a consistent multiplying factor between them, we're often looking at a geometric series. It's a series where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio. In mathematical terms, a geometric series is expressed as the sum of a geometric sequence.

• The formula for the sum of the first n terms of a geometric series is: \[ S_n = \frac{a_1(1 - r^n)}{1 - r} \]
• Here, \( a_1 \) is the first term, r is the common ratio, and n is the number of terms.

With this formula, one can determine finite sums easily, and with a bit of manipulation, this very concept extends to understanding the behavior of sequences as they grow infinitely long, which comes in handy in our exercise to express the sum of exponents as a limit.

Infinite Series

Going Beyond the Finite with Infinite Series

An infinite series carries on forever and is the sum of an infinite sequence of terms. When dealing with infinite series, we're often concerned with their sum, if it exists. Not all infinite series have a finite sum, but when they do, it means the series converges to a specific number.

Geometric series can also be infinite if the number of terms n approaches infinity. The sum of an infinite geometric series with a common ratio r, where \( |r| < 1 \), is given by the formula: \[ S = \frac{a_1}{1 - r} \]

In our exercise, identifying the pattern of the sequence allowed us to utilize this concept of an infinite geometric series. As n increases indefinitely, the term \( \frac{1}{2^n} \) in the exponent becomes negligible, leading the entire sequence to converge to the sum of an infinite series, which in this case, equals 2. This conclusion allows us to say that the infinite series converges, and to determine its sum, illustrating a captivating intersection between abstract patterns and concrete numerical limits.