Sum of Geometric Sequence/Examples/Index to Minus 1
Jump to navigation
Jump to search
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:
- $\ds \sum_{j \mathop = 0}^n x^j = \frac {x^{n + 1} - 1} {x - 1}$
still holds when $n = -1$:
- $\ds \sum_{j \mathop = 0}^{-1} x^j = \frac {x^0 - 1} {x - 1}$
Proof
The summation on the left hand side is vacuous:
- $\ds \sum_{j \mathop = 0}^{-1} x^j = 0$
while on the right hand side we have:
\(\ds \frac {x^{\paren {-1} + 1} - 1} {x - 1}\) | \(=\) | \(\ds \frac {x^0 - 1} {x - 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 0 {x - 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 0\) |
as long as $x \ne 1$.
However, the theorem itself is based on the assumption that $n \ge 0$, so while the result is correct, the derivation to achieve it is not.
$\blacksquare$
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 $9$