Sequence Convergence

Let's analyze the given sequence \(a_n: -16, 12, -9, \frac{27}{4}, -\frac{81}{16}, \ldots\). (a) Finding a formula for \(a_n\): The given sequence seems to follow a geometric progression, where each term is obtained by multiplying the previous term by a constant ratio. In this case, it appears that the common ratio is \(-\frac{3}{4}\) because: \[ \begin{align*} a_2 &= a_1 \times \left(-\frac{3}{4}\right) = -16 \times \left(-\frac{3}{4}\right) = 12 \\ a_3 &= a_2 \times \left(-\frac{3}{4}\right) = 12 \times \left(-\frac{3}{4}\right) = -9 \\ \text{and so on.} \end{align*} \] Therefore, the formula for the \(n\)th term, \(a_n\), is given by: \[ a_n = (-16) \times \left(-\frac{3}{4}\right)^{n-1} \] (b) Finding \( \lim_{n \to \infty} a_n \): To find the limit as \(n\) approaches infinity, we can examine the behavior of the sequence as \(n\) gets larger. In this case, as \(n\) approaches infinity, the term \(\left(-\frac{3}{4}\right)^{n-1}\) will approach zero. Therefore: \[ \lim_{n \to \infty} a_n = (-16) \times 0 = 0 \] (c) Convergence of the series: The series \(\sum_{n=1}^{\infty} a_n\) converges if the limit of the sequence \(a_n\) is zero. In this case, since \(\lim_{n \to \infty} a_n = 0\), the series converges. If a series converges, the sum can be found using the formula for the sum of an infinite geometric series: \[ S = \frac{a_1}{1 - r} \] where \(a_1\) is the first term and \(r\) is the common ratio. In this case: \[ S = \frac{-16}{1 - \left(-\frac{3}{4}\right)} =\frac{-64}{7}\] You can simplify this expression to find the sum.