Question

find the first four terms of the sequence given by the following.
$a_{n} = \frac{(-1)^{n + 1}}{n + 1}$, $n = 1, 2, 3, \dots$

Explanation:

Step1: Find \(a_1\)

Substitute \(n = 1\) into \(a_n=\frac{(-1)^{n + 1}}{n + 1}\).
\(a_1=\frac{(-1)^{1+1}}{1 + 1}=\frac{(-1)^2}{2}=\frac{1}{2}\)

Step2: Find \(a_2\)

Substitute \(n = 2\) into \(a_n=\frac{(-1)^{n + 1}}{n + 1}\).
\(a_2=\frac{(-1)^{2+1}}{2 + 1}=\frac{(-1)^3}{3}=-\frac{1}{3}\)

Step3: Find \(a_3\)

Substitute \(n = 3\) into \(a_n=\frac{(-1)^{n + 1}}{n + 1}\).
\(a_3=\frac{(-1)^{3+1}}{3 + 1}=\frac{(-1)^4}{4}=\frac{1}{4}\)

Step4: Find \(a_4\)

Substitute \(n = 4\) into \(a_n=\frac{(-1)^{n + 1}}{n + 1}\).
\(a_4=\frac{(-1)^{4+1}}{4 + 1}=\frac{(-1)^5}{5}=-\frac{1}{5}\)

Answer:

\(\frac{1}{2},-\frac{1}{3},\frac{1}{4},-\frac{1}{5}\)