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(A) Given $a_{1} = -1$ and $a_{n} = \frac{a_{n-1}}{n}$ for $n \geq 2$. For $n = 2$: $a_{2} = \frac{a_{1}}{2} = \frac{-1}{2}$. For $n = 3$: $a_{3} = \f

Vedclass · Accepted Answer

$(-1) + (\frac{-1}{2}) + (\frac{-1}{6}) + (\frac{-1}{24}) + (\frac{-1}{120}) + \dots$

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(A) Given $a_{1} = -1$ and $a_{n} = \frac{a_{n-1}}{n}$ for $n \geq 2$. For $n = 2$: $a_{2} = \frac{a_{1}}{2} = \frac{-1}{2}$. For $n = 3$: $a_{3} = \frac{a_{2}}{3} = \frac{-1/2}{3} = \frac{-1}{6}$. For $n = 4$: $a_{4} = \frac{a_{3}}{4} = \frac{-1/6}{4} = \frac{-1}{24}$. For $n = 5$: $a_{5} = \frac{a_{4}}{5} = \frac{-1/24}{5} = \frac{-1}{120}$. The first five terms are $-1, \frac{-1}{2}, \frac{-1}{6}, \frac{-1}{24}, \frac{-1}{120}$. The corresponding series is $(-1) + (\frac{-1}{2}) + (\frac{-1}{6}) + (\frac{-1}{24}) + (\frac{-1}{120}) + \dots$

Write the first five terms of the sequence defined by $a_{1} = -1$ and $a_{n} = \frac{a_{n-1}}{n}$ for $n \geq 2$,and obtain the corresponding series.

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