Toolkit 4

Geometric Sequence and Series

an=a1rn1a_n=a_1r^{n-1}
an2=an1an+1a_n^2=a_{n-1}a_{n+1}
Sn=a1+a1r+a1r2++a1rn1=a1(rn1)r1S_n=a_1+a_1r+a_1r^2+\cdots+a_1r^{n-1}=\frac{a_1(r^n-1)}{r-1}
r<1:S=a1+a1r+a1r2+=a11r|r|<1:\quad S_\infty=a_1+a_1r+a_1r^2+\cdots=\frac{a_1}{1-r}

Proof

Let

S=1+r+r2++rn.S = 1 + r + r^2 + \cdots + r^n.

Multiplying by rr, we get

rS=r+r2+r3++rn+1.rS = r + r^2 + r^3 + \cdots + r^{n+1}.

Subtracting the first equation from the second gives

(r1)S=rn+11.(r-1)S = r^{n+1} - 1.

Therefore, for r1r \ne 1,

S=rn+11r1.S = \frac{r^{n+1} - 1}{r - 1}.

If r<1|r| < 1, then rn+10r^{n+1} \to 0 as nn \to \infty. Hence,

1+r+r2+=limn1rn+11r=11r.1 + r + r^2 + \cdots = \lim_{n \to \infty} \frac{1 - r^{n+1}}{1 - r} = \frac{1}{1 - r}. \quad\square