Geometric Progression, Series & Sums

Note: the common ratio of this series is \(-3\), and a series only has a sum to infinity when \(\lvert r \rvert \lt 1\). Since \(\lvert -3 \rvert \gt 1\), this series diverges — it has no sum to infinity, and the formula \(S_\infty = \frac{a_1}{1-r}\) does not apply. The working below shows the calculation the formula produces if it is applied regardless; the result is not a true sum.

1. 1
   Expand the given sum

   \[ \sum_{r=0}^{\infty} 2(-3)^{r+2} = 2(-3)^2 + 2(-3)^3 + 2(-3)^4 + \cdots + 2(-3)^n + \cdots \]

   \[ = 18 - 54 + 162 + \cdots + 2(-3)^n + \cdots \]
2. 2
   Solve for \(r\) using \(r = \frac{a_2}{a_1}\)

   \[ r = \frac{a_2}{a_1} = \frac{-54}{18} = -3 \]
3. 3
   Substitute \(a_1 = 18\) and \(r = -3\) into \(S_\infty = \frac{a_1}{1-r}\)

   \[ S_\infty = \frac{18}{1-(-3)} \]
4. 4
   Subtract \(-3\) from \(1\)

   \[ = \frac{18}{4} \]
5. Simplify \(\frac{18}{4}\)

   \[ S_\infty = \frac{9}{2} \]

   The formula gives nine over two — but see the note above: the series diverges, so it has no sum to infinity.