Question

Test the following series for convergence. $$\sum_{n=1}^{\infty} \frac{(-2)^{n}}{n^{2}}$$

Short answer

The series diverges.

Step-by-step solution

Identify the series

The given series is \(\frac{(-2)^{n}}{n^{2}}\). This series can be identified as an alternating series because of the term \((-2)^{n}\) which alternates in sign.

Apply the Alternating Series Test

To use the Alternating Series Test, two conditions need to be satisfied:1. The absolute value of the terms \(\frac{2^{n}}{n^{2}}\) must be monotonically decreasing2. \(\frac{2^{n}}{n^{2}} \to 0\) as \( n \to fty\)

Check if \( \frac{2^{n}}{n^{2}} \) is decreasing

Examine if \( \frac{2^n}{n^2} \) is monotonically decreasing:Calculate for some values of \( n \) to observe the trend.For \( n = 1, \frac{2^1}{1^2} = 2 \)For \( n = 2, \frac{2^2}{2^2} = 1 \)For \( n = 3, \frac{2^3}{3^2} ≈ 0.89 \)The terms are indeed decreasing.

Check the limit condition

Calculate the limit of \( \frac{2^n}{n^2} \) as \( n \to fty \):due to exponential term in the numerator and square term in denominator: \( \frac{2^n}{n^2} rightarrow 0 \)

Conclude the test

Since one of the Alternating Series Test conditions does not hold (the terms do not approach zero), the series \(∑_{n=1}^{∞} \frac{(-2)^n}{n^2}\) diverges.

Key concepts

Alternating series

An alternating series is a series in which the terms alternately take positive and negative signs. This is usually due to a \rfactor like \r \r \((-1)^n\) or similar. \r \rCommon examples include \r\the alternating harmonic series. To test for convergence in alternating series, we can use the \rAlternating Series Test. \r \rThe test requires two conditions: \r

\r
• The absolute value of the terms must be monotonically decreasing.
\r
• The limit of the terms as \( n \) approaches infinity must be zero.
\r

\r \rRemember: If the series satisfies both conditions, it converges.

Decreasing sequence

A sequence is said to be decreasing if each term is less than or equal to the previous term. This is an important condition for the \rAlternate Series Test. \rTo check if a sequence is decreasing: \r

\r
• Calculate the value of a few terms and observe the trend.
\r
• If each subsequent term is less than or equal to the previous term, the sequence is decreasing.
\r

\rIn the given example, \rwe observe the terms: \r

\r
• For \( n = 1 \), \( \frac{2^1}{1^2} = 2 \)
\r
• For \( n = 2 \), \( \frac{2^2}{2^2} = 1 \)
\r
• For \( n = 3 \), \( \frac{2^3}{3^2} \approx 0.89 \)
\r

\rThis confirms that the terms are indeed decreasing.

Limit comparison test

The Limit Comparison Test is another method to test the convergence of series. It is used when two series are similar in terms of growth rates.\r \rTo use this test: \r

\r
• Choose a comparison series \(b_n\) that resembles the original series \(a_n\).
\r
• Calculate the limit \( \lim_{{n \to \infty}} \frac{{a_n}}{{b_n}} \)
\r

\rIf the limit is a positive finite number, then both series either converge or diverge together. For the series \( \sum \frac{{(-2)^n}}{{n^2}} \), we would typically compare with a known series whose behavior we understand.

Divergence

A series diverges if the terms do not approach zero as \( n \) approaches infinity. \rIn the context of our example, although \rthe terms initially decrease, the crucial condition that \( \frac{{2^n}}{{n^2}} \to 0 \) does not hold as \( n \) approaches infinity due to the exponential growth of the \rnumerator compared to the square \rroot growth of the denominator. This means the series: \r\( \sum_{n=1}^{\infty} \frac{{(-2)^n}}{{n^2}} \) diverges according to the \rAlternating Series Test.