Sum of Geometric Sequence/Examples/Index to Minus 2
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Theorem
Let $x$ be an element of one of the standard number fields: $\Q, \R, \C$ such that $x \ne 1$.
Then the formula for Sum of Geometric Sequence:
- $\ds \sum_{j \mathop = 0}^n x^j = \frac {x^{n + 1} - 1} {x - 1}$
breaks down when $n = -2$:
- $\ds \sum_{j \mathop = 0}^{-2} x^j \ne \frac {x^{-1} - 1} {x - 1}$
Proof
The summation on the left hand side is vacuous:
- $\ds \sum_{j \mathop = 0}^{-2} x^j = 0$
while on the right hand side we have:
\(\ds \frac {x^{\paren {-2} + 1} - 1} {x - 1}\) | \(=\) | \(\ds \frac {x^{-1} - 1} {x - 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {1 / x - 1} {x - 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {1 - x} / x} {x - 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {1 - x} {x \paren {x - 1} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -\frac 1 x\) |
$\blacksquare$
Sources
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.3$: Sums and Products: Exercise $10$