Summing a geometric series

The formula for the sum of a geometric series in its two forms, sums with negative and fractional ratios, sums in sigma notation, and using logarithms to find how many terms exceed a given sum.

The formula

Sum of a GP (r ≠ 1)

$$S_n = \frac{a(r^n - 1)}{r - 1} \qquad\qquad S_n = \frac{a(1 - r^n)}{1 - r}$$

Both forms are the same. Use the first when $r > 1$ and the second when $r < 1$, so the denominator is positive. If $r = 1$, every term is $a$ and $S_n = na$.

Example 1●

Find the sum of the first $10$ terms of $3 + 6 + 12 + \cdots$

$a = 3$, $r = 2$: $\;S_{10} = \dfrac{3(2^{10} - 1)}{2 - 1} = 3 \times 1023 = 3069$.

Example 2: a negative fractional ratio●●

Find the sum of the first $6$ terms of $24 - 12 + 6 - \cdots$

$a = 24$, $r = -\dfrac12$, so $1 - r = \dfrac32$ and $r^6 = \dfrac{1}{64}$.

$S_6 = \dfrac{24\left(1 - \frac{1}{64}\right)}{\frac32} = 24 \times \dfrac{63}{64} \times \dfrac23 = \dfrac{63}{4}$

Example 3: sigma notation●●

Evaluate $\displaystyle\sum_{k=1}^{8} 5 \times 2^{k-1}$.

This is a GP with $a = 5$, $r = 2$ and $8$ terms: $\;5(2^8 - 1) = 1275$.

How many terms to pass a value?

Example 4●●●

How many terms of $2 + 6 + 18 + \cdots$ are needed for the sum to exceed $1\,000\,000$?

$S_n = \dfrac{2(3^n - 1)}{2} = 3^n - 1$.

$3^n - 1 > 1\,000\,000$, so $n > \dfrac{\ln 1\,000\,001}{\ln 3} \approx 12.58$.

$13$ terms are needed. Check: $3^{12} - 1 = 531\,440$ and $3^{13} - 1 = 1\,594\,322$.

Flashcards

Sum of a GP, $r > 1$

$S_n = \dfrac{a(r^n - 1)}{r - 1}$

Sum of a GP, $r < 1$

$S_n = \dfrac{a(1 - r^n)}{1 - r}$

$r = 1$

$S_n = na$

Practice

Find the sum of the first $10$ terms of $3 + 6 + 12 + \cdots$
Find the sum of the first $6$ terms of $24 - 12 + 6 - \cdots$
Evaluate $\displaystyle\sum_{k=1}^{8} 5 \times 2^{k-1}$.
How many terms of $2 + 6 + 18 + \cdots$ are needed for the sum to exceed $1\,000\,000$?
Find $1 + 2 + 4 + \cdots + 512$.
Find $1 - 3 + 9 - 27 + 81$.
Find the sum of the first $6$ terms of $5 + 10 + 20 + \cdots$
When $r = 1$, the sum of the first $n$ terms of a GP is:
For $r = 0.6$, which form keeps the denominator positive?