geometric series

Geometric series

A series where each term is a fixed multiple $q$ of the previous one:

$$\sum _{k=0}^{\infty }{q}^k=1+q+q^2+q^3+\cdots$$

Converges iff $|q|<1$. Diverges for $|q|\ge 1$.

Formulas

For $|q|<1$:

$$\sum _{k=0}^{\infty }{q}^k=\frac{1}{1-q},\sum _{k=1}^{\infty }{q}^k=\frac{q}{1-q}$$

More generally, starting from $k_0$:

$$\sum _{k=k_0}^{\infty }{q}^k=\frac{q^{k_0}-q^{n+1}}{1-q}$$

For infinite sums, the $q^{n+1}$ term goes to zero.

Where does $\frac{1}{1-q}$ come from?

We want a closed formula for $s_n={\sum }_{k=0}^n{q}^k$. The trick: multiplying $s_n$ by $q$ shifts every term by one power, producing an almost-identical sum. Subtracting the two via the telescoping trick will cancel almost everything, leaving an equation we can solve for $s_n$.

$$\begin{array}{rl}{s}_n& =\sum _{k=0}^n{q}^k\\ q\cdot s_n& =\sum _{k=0}^n{q}^{k+1}=\sum _{k=1}^{n+1}{q}^k\end{array}$$

Subtract:

$$\begin{array}{rl}{s}_n-q\cdot s_n& =\sum _{k=0}^n{q}^k-\sum _{k=1}^{n+1}{q}^k\\ & =q^0+\cancel{\sum _{k=1}^n{q}^k}-\cancel{\sum _{k=1}^n{q}^k}-q^{n+1}\\ s_n(1-q)& =1-q^{n+1}\\ s_n& =\frac{1-q^{n+1}}{1-q}(q\ne 1)\\ \stackrel{n\to \infty }{\to }{s}_n& \to \frac{1}{1-q}\text{if }|q|<1\text{, since }{q}^{n+1}\to 0\end{array}$$

If $|q|\ge 1$, then $q^{n+1}$ doesn’t go to zero, so the series diverges.

Example

${\sum }_{k=0}^{\infty }{\left(\frac{1}{2}\right)}^k=\frac{1}{1-1/2}=2$

${\sum }_{k=1}^{\infty }{\left(\frac{1}{5}\right)}^k=\frac{1/5}{1-1/5}=\frac{1/5}{4/5}$
multiply top and bottom by 5: $=\frac{1}{4}$

${\sum }_{k=7}^{\infty }{\left(\frac{1}{2}\right)}^k=\frac{(1/2{)}^7}{1-1/2}=\frac{1/128}{1/2}=\frac{1}{64}$

${\sum }_{k=0}^{\infty }{\left(\frac{41}{42}\right)}^k=\frac{1}{1-41/42}=\frac{1}{1/42}=42$

${\sum }_{k=0}^{\infty }\frac{(-5{)}^k}{4^{2k}}={\sum }_{k=0}^{\infty }{\left(\frac{-5}{16}\right)}^k=\frac{1}{1+5/16}$
multiply top and bottom by 16: $=\frac{16}{21}$