Question

Use the ratio test to determine whether \(\sum_{n=1}^{\infty} a_{n}\) converges, or state if the ratio test is inconclusive. $$ \sum_{n=1}^{\infty} \frac{3^{n^{2}}}{2^{n^{3}}} $$

Short answer

The series converges by the Ratio Test.

Step-by-step solution

Identify the General Term

The given series is \( \sum_{n=1}^{\infty} \frac{3^{n^{2}}}{2^{n^{3}}} \). The general term of the series is \( a_n = \frac{3^{n^{2}}}{2^{n^{3}}} \).

Set Up the Ratio Test

The ratio test requires us to find \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). Compute \( a_{n+1} = \frac{3^{(n+1)^{2}}}{2^{(n+1)^{3}}} \).

Form the Ratio

Compute \( \frac{a_{n+1}}{a_n} = \frac{\frac{3^{(n+1)^{2}}}{2^{(n+1)^{3}}}}{\frac{3^{n^{2}}}{2^{n^{3}}}} = \frac{3^{(n+1)^{2}} \cdot 2^{n^{3}}}{3^{n^{2}} \cdot 2^{(n+1)^{3}}} \). Simplify this expression as much as possible.

Simplify the Powers

Start by expanding the exponents:\[ 3^{(n+1)^{2}} = 3^{n^2 + 2n + 1} = 3^{n^2} \cdot 3^{2n} \cdot 3 \, \] and \[ 2^{(n+1)^3} = 2^{n^3 + 3n^2 + 3n + 1} = 2^{n^3} \cdot 2^{3n^2} \cdot 2^{3n} \cdot 2 \, \] Substitute back to get \( \frac{3^{(n+1)^{2}} \cdot 2^{n^{3}}}{3^{n^{2}} \cdot 2^{(n+1)^{3}}} = \frac{3^{2n+1}}{2^{3n^2 + 3n + 1}} \).

Take the Limit

Now consider the expression \( \left| \frac{3^{2n+1}}{2^{3n^2 + 3n + 1}} \right| \). Calculate:\[ \lim_{n \to \infty} \left| \frac{3^{2n+1}}{2^{3n^2 + 3n + 1}} \right| = \lim_{n \to \infty} \left| \frac{3^{2n+1}}{2^{3n^2 + 3n + 1}} \right| = \lim_{n \to \infty} \frac{3^{2n+1}}{2^{3n^2+3n+1}} \] As \( n \to \infty \), the term with the highest power in the denominator dominates, leading to \( 0 \).

Conclude Based on the Ratio Test

Since \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = 0 \). Since \( L < 1 \), the series \( \sum_{n=1}^{\infty} \frac{3^{n^{2}}}{2^{n^{3}}} \) converges by the Ratio Test.

Key concepts

Convergence of Series

Convergence of a series refers to the behavior of an infinite series as the number of terms grows infinitely large. Specifically, a series is considered convergent if the sum approaches a specific finite number. When dealing with convergence, we're often interested in knowing whether adding together infinite terms will lead to a finite sum or not.

The Ratio Test is a helpful tool in determining convergence. It involves comparing the ratio of consecutive terms in the series. For a series \(\sum_{n=1}^{\infty}a_n\), we find \(L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\). If \(L < 1\), the series is convergent. If \(L > 1\), the series diverges. If \(L = 1\), the test is inconclusive. In our scenario, since \(L = 0\) which is less than 1, the series \(\sum_{n=1}^{\infty} \frac{3^{n^{2}}}{2^{n^{3}}}\) converges.

Infinite Series

An infinite series is the sum of infinitely many terms. Each term adds a bit more to the total, and the series can either converge to a finite number or diverge indefinitely. It's important to investigate the behavior of such series to understand whether their summation leads to meaningful results.

The series \(\sum_{n=1}^{\infty} \frac{3^{n^{2}}}{2^{n^{3}}}\) is one such example. Each term is expressed as a fraction, where the numerator increases due to the power of 3, and the denominator grows even faster due to the power of 2. When analyzing infinite series, the power of the terms and their growth rates can reveal a lot about their convergence or divergence characteristics.

• Faster growing denominators often lead to convergence, as seen in the example.
• Understanding the rate of growth is crucial for determining the behavior of the series.

Calculus

Calculus is a branch of mathematics dealing with rates of change and accumulation. It plays a crucial role in analyzing infinite series. In the context of convergence, calculus provides various tests, such as the Ratio Test, which help determine whether series accumulate to a finite value.

Using calculus in series convergence involves evaluating limits, simplifying terms, and examining how rapidly sequences increase or decrease. In our example, we relied on the limit of ratios of consecutive terms to establish convergence. Calculus allows us to grasp the concept of adding infinitely many terms and understanding their cumulative effect.

Moreover, calculus aids in breaking down complex series into manageable components, making it easier to analyze and derive conclusions about their convergence or divergence.