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Exercise 8.1 · Q12

Q.Write the first five terms of the following sequence and obtain the corresponding series: a1=−1a_1 = -1, an=an−1na_n = \dfrac{a_{n-1}}{n}, n≥2n \geq 2.

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Each term is the previous term divided by the current position number. The first five terms are −1,−12,−16,−124,−1120-1, -\frac{1}{2}, -\frac{1}{6}, -\frac{1}{24}, -\frac{1}{120}, and the series is −1−12−16−124−1120−⋯-1 - \frac{1}{2} - \frac{1}{6} - \frac{1}{24} - \frac{1}{120} - \cdots

Understanding the Recursive Definition

A recursive sequence defines each term using previous terms. Here we're told that a1=−1a_1 = -1 (the seed value) and every subsequent term follows the rule an=an−1na_n = \frac{a_{n-1}}{n} for n≥2n \geq 2. This means to find any term, we divide the previous term by the current position number.

The pattern will emerge quickly: each division by an increasing integer makes the terms shrink in absolute value, and since we start negative, all terms remain negative.

Computing the Terms

  1. First term: Given directly as a1=−1a_1 = -1.

  2. Second term: Apply the recurrence with n=2n = 2:

a2=a12=−12=−12a_2 = \frac{a_1}{2} = \frac{-1}{2} = -\frac{1}{2}

  1. Third term: Now with n=3n = 3:

a3=a23=−123=−16a_3 = \frac{a_2}{3} = \frac{-\frac{1}{2}}{3} = -\frac{1}{6}

  1. Fourth term: With n=4n = 4:

a4=a34=−164=−124a_4 = \frac{a_3}{4} = \frac{-\frac{1}{6}}{4} = -\frac{1}{24}

  1. Fifth term: With n=5n = 5:

a5=a45=−1245=−1120a_5 = \frac{a_4}{5} = \frac{-\frac{1}{24}}{5} = -\frac{1}{120}

Tip

Notice the denominators: 1,2,6,24,1201, 2, 6, 24, 120. These are factorials! We have an=−1n!a_n = -\frac{1}{n!} for all n≥1n \geq 1. You can verify this pattern by induction: if an−1=−1(n−1)!a_{n-1} = -\frac{1}{(n-1)!}, then an=an−1n=−1(n−1)!n=−1n!a_n = \frac{a_{n-1}}{n} = \frac{-\frac{1}{(n-1)!}}{n} = -\frac{1}{n!}.

The Corresponding Series

A series is formed by adding the terms of a sequence. The series corresponding to our sequence is: …

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