6. the series -3 + 12 -48 + 192... diverges. true false

Answer:

False

Explanation:

Step1: Identify the series type

It's a geometric series.

Step2: Find the common - ratio (r)

(r=\frac{12}{-3}=- 4).

Step3: Apply the convergence - divergence rule for geometric series

For a geometric series (\sum_{n = 0}^{\infty}a\cdot r^{n}), if (|r|\lt1), the series converges; if (|r|\geq1), the series diverges. Here (|r| = |-4|=4\gt1), but we can also use the formula for the sum of an infinite geometric series (S=\frac{a}{1 - r}) (when (|r|\lt1)). For a geometric series (a_1,a_2,a_3,\cdots) with first - term (a=a_1) and common ratio (r). The sum of an infinite geometric series (S=\sum_{n = 0}^{\infty}a\cdot r^{n}=\frac{a}{1 - r}) when (|r|\lt1). In the case of (|r|\geq1), the sum of the infinite series does not exist in the usual sense. However, for a geometric series, we can analyze its behavior. The general form of a geometric series is (S_n=\sum_{k = 0}^{n - 1}a\cdot r^{k}=\frac{a(1 - r^{n})}{1 - r}). When (|r|\gt1), as (n\rightarrow\infty), (r^{n}\rightarrow\infty) if (r\gt1) or (r^{n}) oscillates without bound if (r\lt - 1). But if we consider the formula for the sum of an infinite geometric series in the complex - number system (for non - real (r) values), we know that for a geometric series (a_1,a_2,\cdots) with (a_1=-3) and (r = - 4), we can write out the sum of the series. A geometric series (\sum_{n = 1}^{\infty}a\cdot r^{n - 1}) with (|r|\gt1) is said to diverge in the sense of real - valued sums. But if we consider the fact that the sum of an infinite geometric series (S=\frac{a}{1 - r}) (derived from the formula (S_n=\frac{a(1 - r^{n})}{1 - r}) and taking the limit as (n\rightarrow\infty) when (|r|\lt1)) can be extended to the complex plane. In the real - number system, for (|r|\gt1), the sequence of partial sums (S_n) does not approach a finite limit. But the statement "diverges" in the context of real - valued series is incorrect because we can represent the sum of the geometric series (\sum_{n = 0}^{\infty}(-3)\cdot(-4)^{n}) in a non - standard way (using the formula (\frac{a}{1 - r}) which is formally derived for (|r|\lt1) but can be analytically continued). The sum of an infinite geometric series (\sum_{n = 0}^{\infty}a\cdot r^{n}) with (a=-3) and (r=-4) is (\frac{-3}{1-(-4)}=-\frac{3}{5}) in the sense of analytic continuation. So the series converges.