Arithmetic and geometric series

Series and partial sums

A series adds up the terms of a sequence. The sum of the first $n$ terms is the partial sum $S_n$. It has the same units as the terms: a sum of dollars is in dollars.

For the terms 3, 7, 11, 15, the partial sums are $S_1 = 3$, $S_2 = 10$, $S_3 = 21$ and $S_4 = 36$. The partial sums form a sequence of their own.

That sequence has a recursive rule (lesson 1.5.1): each partial sum is the one before plus the next term. Running it backwards recovers any term from two partial sums.

Arithmetic series

Add the whole numbers from 1 to 100. Pair the first with the last, $1 + 100$, then the second with the second-last, $2 + 99$. Every pair makes 101, and there are 50 pairs, so the total is 5050.

The pairing works for every arithmetic sequence, because each step up from the front matches a step down from the back. Write the sum twice, once backwards, and every column holds the first term plus the last.

Before using the formula, count the terms. From $a_1$ to $a_n$ in steps of $d$ there are $\frac{a_n - a_1}{d} + 1$ terms. So $3 + 7 + \cdots + 79$ has $\frac{79 - 3}{4} + 1 = 20$ terms, and its sum is $\frac{20 \times (3 + 79)}{2} = 820$.

A table and the algebra

For 3, 7, 11, …, the $n$-th term is $4n - 1$, so the formula gives $S_n = \frac{n(3 + 4n - 1)}{2} = n(2n + 1)$.

A negative difference

The formula does not care about signs. $40 + 34 + 28 + \cdots + (-20)$ has $\frac{-20 - 40}{-6} + 1 = 11$ terms and sums to $\frac{11 \times (40 - 20)}{2} = 110$. When the first and last terms cancel, as in $15 + 9 + 3 - 3 - 9 - 15$, the sum is 0.

Geometric series

Take $S = 1 + 2 + 4 + 8 + 16$, which is 31. Doubling every term gives $2S = 2 + 4 + 8 + 16 + 32$. The two lists share four terms, so subtracting one from the other leaves only the ends.

The same shift works for any ratio $r$. With first term $a$ and $n$ terms, $rS$ and $S$ share all but two terms, so $S - rS = a - ar^n$. Dividing by $1 - r$ gives the formula.

Lesson PG3.2 in the Proof Gym proves it for every $n$, and shows why it needs $r \ne 1$.

For $5 + 10 + 20 + \cdots$ with 10 terms, $a = 5$ and $r = 2$, so $S_{10} = 5 \times \frac{2^{10} - 1}{2 - 1} = 5115$. For $\frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{1024}$, also 10 terms, the sum is $\frac{1}{2} \times \frac{1 - (1/2)^{10}}{1 - 1/2} = \frac{1023}{1024}$, just under 1.

A table and the algebra

For $3 + 6 + 12 + \cdots$, $a = 3$ and $r = 2$, so the formula gives $S_n = 3(2^n - 1)$.

With ratio 2, each new term beats all the earlier terms put together: the fourth term, 24, is more than $S_3 = 21$. That is why doubling sums run away.

Every ratio

• $r$ above 1. The terms grow and the sums run away, as 5115 and 189 above.
• $r$ between 0 and 1. The terms shrink and the partial sums creep up towards a ceiling, as $\frac{1023}{1024}$ above. Lesson 1.5.3 finds the ceiling.
• $r$ negative. The terms alternate, and so do the steps of the partial sums. For $1 - 2 + 4 - 8 + 16$, the partial sums are $1, -1, 3, -5, 11$. The formula still works, with brackets: $\frac{1 - (-2)^5}{1 - (-2)} = \frac{33}{3} = 11$.
• $r = -1$. $1 - 1 + 1 - 1 + \cdots$ has partial sums $1, 0, 1, 0$. The formula gives $\frac{1 - (-1)^n}{2}$: 1 for odd $n$, 0 for even $n$.
• $r = 1$. The formula would divide by zero. Every term is $a$, so $S_n = na$: the sum of a constant from lesson 0.2.2.

Reading a sum in $\Sigma$ notation

A sum in $\Sigma$ notation (lesson 0.2.2) hides the three numbers the geometric formula needs. Read them off before you use it.

1. The first term: put the lowest index into the term.
2. The ratio: what each step multiplies by.
3. The number of terms: top limit minus bottom limit, plus 1.

$\sum_{k=1}^{6} 2 \cdot 3^k$ starts at $k = 1$, so its first term is $2 \times 3 = 6$, not 2. The ratio is 3 and there are $6 - 1 + 1 = 6$ terms, so the sum is $6 \times \frac{3^6 - 1}{3 - 1} = 2184$.

$\sum_{k=3}^{10} \left(\frac{1}{2}\right)^k$ has first term $\frac{1}{8}$, ratio $\frac{1}{2}$ and $10 - 3 + 1 = 8$ terms. Its sum is $\frac{1}{8} \times \frac{1 - (1/2)^8}{1 - 1/2} = \frac{255}{1024}$, about 0.249.

The same reading works for arithmetic sums: $\sum_{k=1}^{20} (4k - 1)$ is $3 + 7 + \cdots + 79$ again, 820.

Lesson 1.5.3 lets the number of terms grow without end, and case study 1.5.5 prices a run of equal payments as a geometric series.

Common mistakes

1. Count the terms. $\sum_{k=m}^{n}$ has $n - m + 1$ terms, and $3 + 7 + \cdots + 79$ has $\frac{79 - 3}{4} + 1 = 20$, not 19.
2. The formula's $a$ is the first term actually added. For $\sum_{k=1}^{6} 2 \cdot 3^k$ it is 6, not 2.
3. Ratio 1 needs its own rule. The geometric formula divides by zero; use $na$.
4. Bracket a negative ratio. $(-2)^5 = -32$, and $1 - (-2) = 3$.
5. Keep the two forms whole. $a\frac{1 - r^n}{1 - r}$ and $a\frac{r^n - 1}{r - 1}$ are equal; $a\frac{1 - r^n}{r - 1}$ has the wrong sign.
6. The arithmetic formula needs the last term. If you only know $d$, use $\frac{n}{2}(2a_1 + (n - 1)d)$.