Geometric series appear frequently in time series analysis, particularly when working with ARMA models and spectral analysis. Understanding their properties is essential for many derivations in this course.
Finite Geometric Series ¶ Let S = x + x 2 + x 3 + … + x N = ∑ n = 1 N x n S = x + x^{2} + x^{3} + \ldots + x^{N} = \sum_{n=1}^{N}x^{n} S = x + x 2 + x 3 + … + x N = ∑ n = 1 N x n geometric progression .
Now consider x S xS x S
x S = x 2 + x 3 + x 4 + … + x N + 1 = ∑ n = 2 N + 1 x n xS = x^{2} + x^{3} + x^{4} + \ldots + x^{N+1} = \sum_{n=2}^{N+1}x^{n} x S = x 2 + x 3 + x 4 + … + x N + 1 = n = 2 ∑ N + 1 x n Subtracting these:
( 1 − x ) S = S − x S = ( x + x 2 + x 3 + … + x N ) − ( x 2 + x 3 + x 4 + … + x N + 1 ) = x − x N + 1 = x ( 1 − x N ) \begin{split}
(1-x)S &= S - xS\\
&= (x + x^{2} + x^{3} + \ldots + x^{N}) - (x^{2} + x^{3} + x^{4} + \ldots + x^{N+1})\\
&= x - x^{N+1}\\
&= x(1-x^{N})
\end{split} ( 1 − x ) S = S − x S = ( x + x 2 + x 3 + … + x N ) − ( x 2 + x 3 + x 4 + … + x N + 1 ) = x − x N + 1 = x ( 1 − x N ) Therefore:
S = x 1 − x N 1 − x for x ≠ 1. S = x\frac{1-x^{N}}{1-x} \quad \text{for } x \neq 1. S = x 1 − x 1 − x N for x = 1. Prove that ∑ n = 0 N x n = 1 − x N + 1 1 − x for x ≠ 0 , 1 \sum_{n=0}^{N}x^{n}=\frac{1-x^{N+1}}{1-x} \quad \text{for } x \neq 0,1 ∑ n = 0 N x n = 1 − x 1 − x N + 1 for x = 0 , 1
Solution: This follow immediately from the derivation in Eq. (2) T = 1 + x + x 2 + … + x N = ∑ n = 0 N x n T = 1 + x + x^2 + \ldots + x^{N} = \sum_{n=0}^{N}x^{n} T = 1 + x + x 2 + … + x N = ∑ n = 0 N x n
( 1 − x ) T = T − x T = ( 1 + x + x 2 + … + x N ) − ( x + x 2 + x 3 + … + x N + 1 ) = 1 − x N + 1 \begin{split}
(1-x)T &= T - xT\\
&= (1+ x + x^{2} + \ldots + x^{N}) - (x + x^{2} + x^{3} + \ldots + x^{N+1})\\
&= 1 - x^{N+1}\\
\end{split} ( 1 − x ) T = T − x T = ( 1 + x + x 2 + … + x N ) − ( x + x 2 + x 3 + … + x N + 1 ) = 1 − x N + 1 Thus we arrive at
T = 1 − x N + 1 1 − x T = \frac{1 - x^{N+1}}{1-x} T = 1 − x 1 − x N + 1 Infinite Geometric Series ¶ Eq. (3) x ≠ 1 x \neq 1 x = 1
Case 1: If ∣ x ∣ > 1 |x| > 1 ∣ x ∣ > 1 N N N
Case 2: For ∣ x ∣ < 1 |x| < 1 ∣ x ∣ < 1 lim N → ∞ x ( 1 − x N ) = x \lim_{N\to\infty}x(1- x^N) = x lim N → ∞ x ( 1 − x N ) = x
lim N → ∞ ∑ n = 1 N x n = x 1 − x for ∣ x ∣ < 1 \lim_{N\to\infty}\sum_{n=1}^{N}x^{n} = \frac{x}{1-x} \quad \text{for } |x| < 1 N → ∞ lim n = 1 ∑ N x n = 1 − x x for ∣ x ∣ < 1 And starting from n = 0 n=0 n = 0 lim N → ∞ ( 1 − x N ) = 1 \lim_{N\to\infty}(1- x^N) = 1 lim N → ∞ ( 1 − x N ) = 1
lim N → ∞ ∑ n = 0 N x n = 1 1 − x for ∣ x ∣ < 1 , x ≠ 0 \lim_{N\to\infty}\sum_{n=0}^{N}x^{n} = \frac{1}{1-x} \quad \text{for } |x|<1, x\neq0 N → ∞ lim n = 0 ∑ N x n = 1 − x 1 for ∣ x ∣ < 1 , x = 0 The value of Eq. (7) x = 0 x=0 x = 0 00 , which in different contexts may be either undefined or defined as 1.
We can derive several related infinite sums from the basic formula:
Alternating Series ¶ ∑ n = 1 ∞ ( − 1 ) n x n = ∑ n = 1 ∞ ( − x ) n = − x 1 − ( − x ) = − x 1 + x \sum_{n=1}^{\infty}(-1)^n x^{n} = \sum_{n=1}^{\infty}(-x)^{n} = \frac{-x}{1-(-x)} = \frac{-x}{1+x} n = 1 ∑ ∞ ( − 1 ) n x n = n = 1 ∑ ∞ ( − x ) n = 1 − ( − x ) − x = 1 + x − x Even Powers ¶ ∑ n = 1 ∞ x 2 n = ∑ n = 1 ∞ ( x 2 ) n = x 2 1 − x 2 \sum_{n=1}^{\infty}x^{2n} = \sum_{n=1}^{\infty}(x^2)^{n} = \frac{x^2}{1-x^2} n = 1 ∑ ∞ x 2 n = n = 1 ∑ ∞ ( x 2 ) n = 1 − x 2 x 2 We will come back to Eq.s (6) (7)