Question

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

Short answer

The series converges by the ratio test.

Step-by-step solution

Identify the terms

The given series is \( \sum_{n=1}^{\infty} a_n \) where \( a_n = \frac{2^{n^2}}{n^n \times n!} \). We will use the ratio test to analyze the convergence.

Apply the ratio test

The ratio test involves finding the limit \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). If \( L < 1 \), the series converges; if \( L > 1 \), it diverges; if \( L = 1 \), the test is inconclusive.

Find \( a_{n+1} \)

Calculate \( a_{n+1} = \frac{2^{(n+1)^2}}{(n+1)^{n+1} \times (n+1)!} \).

Compute \( \frac{a_{n+1}}{a_n} \)

Find the ratio \( \frac{a_{n+1}}{a_n} = \frac{2^{(n+1)^2}}{(n+1)^{n+1} \times (n+1)!} \times \frac{n^n \times n!}{2^{n^2}} \).

Simplify the expression

Simplify the fraction: \( \frac{a_{n+1}}{a_n} = \frac{2^{2n+1} \times n^n \times n!}{(n+1)^{n+1} \times (n+1)!} \).

Evaluate the limit

Evaluate the limit \( L = \lim_{n \to \infty} \left( \frac{2^{2n+1} \times n^n \times n!}{(n+1)^{n+1} \times (n+1)!} \right) \).

Analyze the limit

The terms \( n^n \) and \( (n+1)^{n+1} \) will result in behavior tending to zero as \( n \) grows because \( (n+1)^{n+1} \) grows faster than \( n^n \). Thus, after canceling appropriate terms and simplifying, \( L = 0 \). Since \( L < 1 \), the series converges.

Key concepts

Series Convergence

Series convergence is a fundamental concept in calculus that helps to understand if an infinite series approaches a particular value, or simply put, settles down to a certain sum. When you deal with infinite series, you're essentially adding up an unlimited number of terms.
This might sound overwhelming, but the idea of series convergence tells us whether these sums actually lead to a finite number or not.

In the context of this problem, we used the ratio test to test for convergence.
The ratio test is especially handy when dealing with series that are complicated or involve factorials or exponential terms.
For our given series, which involves exponential expressions and factorials, identifying whether it converges or not can be efficiently managed by the ratio test. Together with other tests such as the root test, it offers a straightforward way to predict this behavior.

Recognizing series convergence is essential in many fields, from physics to finance, where modeling relies on predicting definite outcomes.

Infinite Series

An infinite series is the sum of infinitely many terms. Unlike finite sums, infinite series goes on without end.
They're written in the form \( \sum_{n=1}^{\infty} a_n \), where \( a_n \) represents the sequence of terms.

One of the interesting aspects of infinite series is whether they add up to a finite number, or diverge to infinity.
This particular property can be fascinating, as not all infinite series behave the same. Some converge, meaning they settle at a certain constant value. Others diverge, endlessly increasing without bounds.

To study the behavior of our given series \( \sum_{n=1}^{\infty} 2^{n^2} / (n^n n!) \), math experts use specialized techniques like the ratio test, to understand if the series converges and what implications it has in various applicable scenarios.
This understanding is pivotal when handling calculations involving series in both pure and applied mathematics.

Calculus Series Testing

Calculus series testing involves using specific tests to determine the convergence of a series.
These tests are integral tools in calculus to analyze the behavior of series under certain conditions.

One such test is the Ratio Test, which is frequently employed when a series includes exponential and factorial characteristics, like the one in our exercise. The idea is to look at the ratio of successive terms in the series, specifically focusing on \[ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \].
Here, \( a_n \) refers to the terms of the series.

If \( L < 1 \), it suggests that each term progressively gets smaller, hinting at convergence.
If \( L > 1 \), the series diverges, because terms tend to grow larger.
When \( L = 1 \), the test becomes inconclusive.
For the series tested here, \( L = 0 \), which clearly tells us the series converges.
This process simplifies the immense complexity often involved when dealing with intricate series, enabling an understanding of their converging behaviors. Other tests include the Root Test, Alternating Series Test, and Integral Test, each providing unique insights into different types of series.