Question

Use the ratio test to find whether the following series converge or diverge: $$\sum_{n=1}^{\infty} \frac{2^{n}}{n^{2}}$$

Short answer

The series diverges.

Step-by-step solution

Identify the general term of the series

The general term of the series is given by \(a_n = \frac{2^n}{n^2}\). We will use this in the ratio test.

Set up the ratio test

The ratio test involves computing the limit of \( \left| \frac{a_{n+1}}{a_n} \right| \) as n approaches infinity.

Compute \(a_{n+1}\)

To find \(a_{n+1}\), substitute \(n + 1\) for \(n\) in the general term: \(a_{n+1} = \frac{2^{n+1}}{(n+1)^2}\).

Form the ratio \( \frac{a_{n+1}}{a_n} \)

Substitute \(a_{n+1}\) and \(a_n\) into the ratio: \[\frac{a_{n+1}}{a_n} = \frac{ \frac{2^{n+1}}{(n+1)^2} }{ \frac{2^n}{n^2} } = \frac{2 \cdot 2^n \cdot n^2}{(n+1)^2 \cdot 2^n} = \frac{2n^2}{(n+1)^2} \]

Simplify the ratio

Simplify \( \frac{2n^2}{(n+1)^2} \): \[ \frac{2n^2}{(n+1)^2} = 2 \cdot \frac{n^2}{(n+1)^2} = 2 \cdot \frac{n^2}{n^2 + 2n + 1} = 2 \cdot \left(\frac{1}{1 + \frac{2}{n} + \frac{1}{n^2}} \right) \]

Take the limit as \(n \to \infty\)

Now, find the limit of \( \left| \frac{a_{n+1}}{a_n} \right| \) as n approaches infinity: \[ \lim_{n \to \infty} 2 \cdot \left( \frac{1}{1 + \frac{2}{n} + \frac{1}{n^2}} \right) = 2 \cdot \frac{1}{1+0+0} = 2 \cdot 1 = 2 \]

Apply the ratio test conclusion

According to the ratio test, if this limit \( L \) is greater than 1, the series diverges. Since \( L = 2 > 1 \), the series \[\sum_{n=1}^{\infty} \frac{2^n}{n^2} \] diverges.

Key concepts

Convergence and Divergence of Series

When we talk about series in mathematics, we often want to know if a series converges or diverges. Convergence means that the sum of the series approaches a specific value as we add more and more terms.
Divergence implies that the sum either increases without bound or does not approach a specific value.
There are various tests to determine convergence or divergence of a series, one of which is the ratio test.
The ratio test is particularly useful for series where each term is multiplied by a constant factor from one term to the next. For instance, the series \( \sum_{n=1}^{\infty} \frac{2^n}{n^2} \) utilizes the ratio test efficiently.
By checking the limit of the ratio of consecutive terms, we can decide if the series converges or diverges.

Limits in Sequences and Series

Understanding limits is essential for analyzing sequences and series. In the context of the ratio test, we calculate the limit of the ratio of consecutive terms as n approaches infinity.
Algebraic manipulation often simplifies this process.
For example, in our series \( \sum_{n=1}^{\infty} \frac{2^n}{n^2} \), we found the ratio between consecutive terms was \( \frac{2n^2}{(n+1)^2} \).
This ratio simplifies to \( 2 \cdot \frac{n^2}{n^2 + 2n + 1} \), which gives us 2 when we take the limit as n approaches infinity.
Therefore, the limit becomes \lim_{n \to \infty} 2 \cdot \frac{1}{1 + \frac{2}{n} + \frac{1}{n^2}} = 2 \cdot 1 = 2\. Knowing this result helps us apply the ratio test conclusion.

Mathematical Analysis

Mathematical analysis involves deep, theoretical methods for solving mathematical problems.
It includes tools and techniques for understanding concepts such as limits, sequences, and series.
In our example, we applied the ratio test, which is a standard method in mathematical analysis for determining series convergence or divergence.
This test provides a conclusion based on the behavior of the sequence of terms in the series as we approach infinity.
By evaluating the limit \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\), if \( L \) is less than 1, the series converges; if \( L \) equals 1, the test is inconclusive; and if \( L \) is greater than 1, the series diverges.
In our case, since \( L \) is 2, which is greater than 1, we conclude that the series \( \sum_{n=1}^{\infty} \frac{2^n}{n^2} \) divergently breaks the criteria for convergence.