Question

Use Theorem 6 to find the limit of the following sequences or state that they diverge. $$\left\\{\frac{n^{1000}}{2^{n}}\right\\}$$

Short answer

Sequence: $$\frac{n^{1000}}{2^{n}}$$ Answer: The limit of the given sequence is 0 as n approaches infinity.

Step-by-step solution

Define the numerator and the denominator sequences

In this case, we have the following sequences: Numerator sequence: \(a(n) = n^{1000}\) Denominator sequence: \(b(n) = 2^{n}\)

Determine the limits of the individual sequences

Now, we need to find the limit of these sequences as n approaches infinity. \(a(n)\): For the numerator sequence, as n approaches infinity, \(n^{1000}\) will also approach infinity. \(b(n)\): For the denominator sequence, as n approaches infinity, \(2^{n}\) will also approach infinity. Since both sequences approach infinity as n approaches infinity, we need to compare their growth rates.

Compare the growth rates

To compare the growth rates, consider the natural logarithm of the sequences: For the numerator sequence: \(\ln{a(n)} = \ln{(n^{1000})} = 1000\ln{n}\) For the denominator sequence: \(\ln{b(n)} = \ln{(2^{n})} = n\ln{2}\) As n approaches infinity, both \(1000\ln n\) and \(n\ln 2\) approach infinity. However, \(n\ln 2\) grows faster than \(1000\ln n\), so the denominator grows faster.

Apply Theorem 6

According to Theorem 6, since the denominator grows faster than the numerator, the limit of the sequence is zero. The limit of the sequence is: $$\lim_{n\to\infty}\frac{n^{1000}}{2^{n}} = 0$$ The given sequence converges to zero as n approaches infinity.

Key concepts

Growth Rates

When analyzing sequences, especially when determining their limits, understanding growth rates is crucial. Growth rates tell us how quickly a sequence increases or decreases as we move towards infinity. In our specific exercise, the focus is on comparing the growth rate of the numerator, \(n^{1000}\), with the denominator, \(2^{n}\).

Notice that even though both sequences move towards infinity, they do so at different speeds. The key is to identify which sequence grows faster. In this context, the denominator \(2^{n}\) increases at an exponential rate, while the numerator \(n^{1000}\) increases polynomially.

• Exponential growth implies multiplying the base repeatedly. So, it grows extremely fast.
• Polynomial growth, on the other hand, means you multiply the variable by itself a fixed number of times.

This distinction is vital since it helps determine the convergence behavior of the whole sequence.

Numerator and Denominator Sequences

The numerator and denominator sequences form the backbone of our problem. By identifying them, we categorize and understand different parts of the sequence we are looking into.

• Numerator sequence: In our sequence \(\frac{n^{1000}}{2^{n}}\), the numerator sequence is given by \(a(n) = n^{1000}\).
• Denominator sequence: The denominator, in this case, is \(b(n) = 2^{n}\).

Recognizing these components separately allows us to analyze their individual behavior as \(n\) approaches infinity. Both approach infinity, but their rate becomes important in the final analysis. Understanding each sequence individually helps us see how they interact when put in a fraction.

Convergence and Divergence

Convergence and divergence are central concepts when working with sequences and series. A sequence converges if it approaches a particular value as \(n\) goes to infinity. On the other hand, divergence occurs if the sequence continues to increase indefinitely or oscillates without settling on a value.

In our sequence \(\frac{n^{1000}}{2^{n}}\), convergence is examined by comparing growth rates. Since \(2^{n}\) grows faster than \(n^{1000}\), the fraction's terms shrink towards zero as \(n\) becomes very large.

• Because the denominator grows faster, \(\lim_{n\to\infty}\frac{n^{1000}}{2^{n}} = 0\).
• Therefore, the sequence converges to zero, indicating it becomes negligibly small as \(n\) increases.

This conclusion is a result of the faster growth rate of the exponential denominator overpowering the polynomial numerator, leading the sequence towards convergence.

Natural Logarithms

Natural logarithms play a pivotal role in comparing the growth rates of sequences. By transforming sequences into their logarithmic form, it's often easier to analyze their behavior as \(n\) approaches infinity.

Considering our problem:

• The natural logarithm of the numerator is \(\ln(n^{1000}) = 1000\ln(n)\).
• The natural logarithm of the denominator is \(\ln(2^{n}) = n\ln(2)\).

This transformation reveals that \(n\ln(2)\) grows faster than \(1000\ln(n)\), because the exponential increase outpaces the polynomial one.

Natural logarithms thus provide a clear analytic pathway, helping us quantify and compare how fast different sequences grow. Applying natural logarithms simplifies our assessment of which part of the sequence grows faster, directly impacting our conclusions about convergence or divergence.