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Mathematics · Ch 5 — Binomial Theorem, Sequences and Series

Infinite Geometric Series

5.6.2

Infinite Geometric Series

Infinite Series. If (an)(a_n) is an infinite sequence, the formal expression a1+a2+⋯a_1+a_2+\cdots is an infinite series, denoted ∑k=1∞ak\displaystyle\sum_{k=1}^\infty a_k. As set up in §5.6, its convergence and sum (when it exists) are governed by the limit of the partial-sum sequence sn=a1+⋯+ans_n=a_1+\cdots+a_n.

The geometric series ∑xn\sum x^n is the prototype. Taking sn=x0+x1+⋯+xn=1−xn+11−xs_n=x^0+x^1+\cdots+x^n=\dfrac{1-x^{n+1}}{1-x} (x≠1x\ne1): since xn→0x^n\to0 exactly when ∣x∣<1|x|<1, we get sn→11−xs_n\to\dfrac1{1-x} under that same condition. So:

∑n=0∞xn=11−x,∣x∣<1,i.e.11−x=1+x+x2+x3+⋯  (∣x∣<1).\sum_{n=0}^\infty x^n = \frac1{1-x}, \quad |x|<1, \qquad\text{i.e.}\qquad \frac1{1-x}=1+x+x^2+x^3+\cdots\ \ (|x|<1).

Three companion series follow by substitution:

∑n=0∞(−1)nxn=11+x, ∣x∣<1 ⟹ 11+x=1−x+x2−x3+⋯\sum_{n=0}^\infty(-1)^nx^n = \frac1{1+x},\ |x|<1 \ \Longrightarrow\ \frac1{1+x}=1-x+x^2-x^3+\cdots

∑n=0∞(2x)n=11−2x, ∣x∣<12 ⟹ 11−2x=1+2x+4x2+8x3+⋯\sum_{n=0}^\infty(2x)^n = \frac1{1-2x},\ |x|<\tfrac12 \ \Longrightarrow\ \frac1{1-2x}=1+2x+4x^2+8x^3+\cdots …