Problem 17 - Thomas Calculus

Question

Determining Convergence or Divergence In Exercises \(17-44,\) use any method to determine if the series converges or diverges. Give reasons for your answer. $$\sum_{n=1}^{\infty} \frac{n^{\sqrt{2}}}{2^{n}}$$

Step-by-Step Solution

Answer

The series converges by the ratio test.

Step 1: Analyze the General Term of the Series

Consider the general term of the series \(a_n = \frac{n^{\sqrt{2}}}{2^n}\). The numerator increases with \(n\), but the denominator \(2^n\) grows exponentially faster because it is a geometric factor.

Step 2: Apply the Ratio Test

The ratio test gives us the limit \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). Calculate \(\frac{a_{n+1}}{a_n} = \frac{(n+1)^{\sqrt{2}}}{2^{n+1}} \times \frac{2^n}{n^{\sqrt{2}}} = \frac{(n+1)^{\sqrt{2}} \cdot 2^n}{2^{n+1} \cdot n^{\sqrt{2}}} = \frac{(n+1)^{\sqrt{2}}}{2 \cdot n^{\sqrt{2}}}. \)Simplifying further, we have:\(\frac{(n+1)^{\sqrt{2}}}{2 \cdot n^{\sqrt{2}}} = \frac{1}{2} \left( 1 + \frac{1}{n} \right)^{\sqrt{2}}.\)

Step 3: Evaluate the Limit

As \(n\) approaches infinity, the term \(\left( 1 + \frac{1}{n} \right)^{\sqrt{2}}\) approaches 1 due to the property of limits. Thus, \[ L = \lim_{n \to \infty} \frac{1}{2} \left( 1 + \frac{1}{n} \right)^{\sqrt{2}} = \frac{1}{2}. \]

Step 4: Determine Convergence or Divergence from the Ratio Test

Since \(L = \frac{1}{2} < 1\), according to the ratio test, the series \(\sum_{n=1}^{\infty} \frac{n^{\sqrt{2}}}{2^n}\) converges.

Key Concepts

Ratio Test

The ratio test is a powerful tool in determining the convergence or divergence of an infinite series. When a series \(\sum_{n=1}^{\infty} a_n\) is presented, the ratio test analyzes the limit of the absolute value of the ratio of consecutive terms. More explicitly, it examines the limit \L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\.

• If \(L < 1\), the series converges.
• If \(L > 1\), or if the limit does not exist, the series diverges.
• If \(L = 1\), the test is inconclusive.

In our example, after simplifying the ratio, we found that \\frac{a_{n+1}}{a_n} = \frac{1}{2} \left( 1 + \frac{1}{n} \right)^{\sqrt{2}}.\
As \(n\) approaches infinity, \\left( 1 + \frac{1}{n} \right)^{\sqrt{2}}\ approaches 1, leading to \(L = \frac{1}{2}\).
Because \(L = \frac{1}{2} < 1\), the series converges according to the ratio test rules.

Infinite Series

An infinite series is simply the sum of an infinite sequence of terms. When dealing with infinite series, we are often concerned with their convergence or divergence.

• If a series converges, the sum of its terms approaches a specific finite number.
• If a series diverges, the sum doesn't approach any finite limit.

In the example provided, the series is \(\sum_{n=1}^{\infty} \frac{n^{\sqrt{2}}}{2^n}\). This means we are adding up an infinite number of terms of the form \(\frac{n^{\sqrt{2}}}{2^n}\). Understanding whether such a series sums to a specific number or tends to infinity is crucial in mathematical analysis and helps in determining the nature and behaviors of functions and models built using that series.

Exponential Growth

Exponential growth refers to quantities growing at a rate proportional to their current value. Imagine something growing faster the bigger it gets. In our series, the denominator \(2^n\) illustrates this concept.
When we compare the growth rates of \(n^{\sqrt{2}}\) (numerator) and \(2^n\), we see that the exponential term, \(2^n\), grows much quicker than the power term \(n^{\sqrt{2}}\).

• The exponential growth in the denominator is critical in determining convergence.
• Even though the numerator grows, the denominator's growth far outpaces it.

This discrepancy in growth rates results in the terms of the series becoming very small as \(n\) increases, which is a significant factor in driving the series towards convergence.