Alternating Series Test

If the terms of the alternating series \(\sum_{n=1}^{\infty}(-1)^{n+1}b_{n}=b_{1}-b_{2}+b_{3}-...\mathrm{~where~}b_{n}>0\) satisfy 

1. \(b_{n+1}\leq b_n\) for all \(n \geq 1\) (\(b_{n}\) is decreasing)
2. \(\lim_{n\to\infty}b_n^{\color{red}{\cdot}}=0\)

then the series is convergent.

Example: 

Determine whether the following series are convergent or divergent.

1. \(\sum_{n=2}^\infty\frac{n(-1)^n}{\ln n}\)
2. \(\sum_{n=2}^\infty\frac{\cos(n\pi)}{n^{3/4}}\)

Solution:

1. Checking condition 2 of the alternative series test.

Let \(b_n=\frac n{\ln n}\). Using L’Hôpital’s rule to compute the limit, 

$$\begin{gathered}
\operatorname*{lim}_{n\to\infty}b_{n} =\lim_{n\to\infty}\frac{n}{\ln n}=\lim_{n\to\infty}\frac{1}{1/n} \\
=\lim_{n\to\infty}n=\infty\neq0 
\end{gathered}$$

Thus, condition 2 is not satisfied, and the alternating series \(\sum_{n=2}^\infty\frac{n(-1)^n}{\ln n}\) is divergent.

2. Note that \(\sum_{n=1}^{\infty}\cos(n\pi)=\sum_{n=1}^{\infty}(-1)^n\).

Then, the series can be written as \(\sum_{n=1}^{\infty}\frac{\cos(n\pi)}{n^{3/4}}=\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n^{3/4}}.\)

Checking both conditions of the alternating series test.

Let \(b_{n}=\frac{1}{n^{3/4}}.\) Clearly \(b_{n}>0.\)

Condition 1:

Let \(f(x)=\frac{1}{x^{3/4}}.\) Then, \(f^{\prime}(x)=-\frac{3}{4x^{7/4}}<0\) for all \(x\) values in \([1,\infty)\). This means that \(b_{n}\) is decreasing for all \(n\) in \([1,\infty)\).

Condition 2:

Remember the fact:

\(\lim_{n\to\infty}\frac1{n^p}=0\) where \(p>0\) is a real number.

\(\lim\limits_{n\to\infty}b_n=\lim\limits_{n\to\infty}\frac{1}{n^{3/4}}=0\).

Both conditions of the alternating series test are satisfied. Thus, \(\sum_{n=1}^\infty\frac{\cos(n\pi)}{n^{3/4}}\) is convergent.