Series & Sigma Notation — Print

Adding a long list of terms without adding them one at a time.

What the practice questions cover

These questions use 5 ideas from the course. Follow any of them for a definition, where it is taught, and what it leads to.

Add every whole number from 1 to 100.

The story goes that a schoolmaster set this to keep a class quiet, and one boy answered in seconds.

add 1 to 100 — without adding 100 things 1 100 101 2 99 101 3 98 101 4 97 101 ⋮⋮ fifty pairs of 101 — the adding is already done

He paired the ends: 1 + 100 = 101, 2 + 99 = 101, 3 + 98 = 101. Every pair makes 101.

There are 50 such pairs, so the total is 50 × 101 = 5050. No adding required.

That trick is the formula for an arithmetic series: pair the ends, count the pairs. Every series formula in this unit is a shortcut of the same kind.

A growing tile pattern with rule 3n plus 2 n = 15n = 28n = 311n = 414 step of 3 each time, plus a starting 2 → rule: 3n + 2

Reading sigma notation

∑k=14 (2k+1) says: let k run from 1 to 4, work out 2k+1 each time, and add the results.

That gives 3 + 5 + 7 + 9 = 24. The notation is an instruction, not a new operation.

The letter underneath is just a counter — renaming it j or i changes nothing.

Read it as a loop

Bottom number is where the counter starts, top is where it stops, and the expression is what to add each pass.

Arithmetic series: pair the ends

When each term adds the same amount, the first and last average out across the whole list.

So Sn = n2(a1 + an) — the number of terms, times the average of the ends.

For 1 to 100 that is 1002(1 + 100) = 5050, exactly Gauss’s pairing.

Count the terms carefully

From 7 to 31 in steps of 3 there are 31-73 + 1 = 9 terms. The +1 is the one people forget — both ends are included.

Geometric series

When each term multiplies by the same ratio, the sum is Sn = a1 1 - rn1 - r.

For r=1, use Sn=na1 instead. For r ≠ 1, it comes from a neat cancellation: subtract r times the series from the series, and almost everything vanishes.

For 2 + 6 + 18 + 54: a1 = 2, r = 3, n = 4, giving 21-811-3 = 80.

Find r by dividing

Divide any term by the one before it. If you get the same number every time it is geometric; if not, check whether it is arithmetic instead.

When an infinite series has a total

If |r| < 1 the terms shrink towards nothing, and the sum settles on S = a11 - r.

That is why 12 + 14 + 18 + ⋯ comes to exactly 1 — it converges.

For a nonzero first term, if |r| ≥ 1 the partial sums do not converge. They may grow without bound or oscillate, as they do when r=-1.

Check the ratio first

The infinite formula only applies when |r| < 1. Using it on r = 2 produces a confident, meaningless number — often a negative one.

Repeating decimals are geometric series

0.7 means 0.7 + 0.07 + 0.007 + ⋯, a geometric series with a1 = 0.7 and r = 0.1.

The formula gives 0.71 - 0.1 = 0.70.9 = 79.

So a repeating decimal is exactly a fraction, not approximately one.

Where the nines come from

The 9 in 79 is 1 - 0.1 in disguise. Two repeating digits give 1 - 0.01, and hence a denominator of 99.

Worked examples

Expanding sigma notation

Evaluate ∑k=15 (3k - 1).

  1. Run k from 1 to 5: 2, 5, 8, 11, 14. Substitute each counter value in turn.
  2. These go up by 3, so it is arithmetic with 5 terms. A constant difference.
  3. Sum = 52(2 + 14) = 52(16) = 40. Number of terms times the average of the ends.

40

A geometric sum

Find the sum of the first 6 terms of 3, 6, 12, …

  1. Divide to find r: 6 ÷ 3 = 2. A constant ratio means geometric.
  2. Apply Sn = a11-rn1-r = 31 - 641 - 2. r6 = 26 = 64.
  3. = 3 · -63-1 = 189. Check by listing: 3+6+12+24+48+96 = 189.

189

An infinite sum

Find the sum of 12 + 4 + 43 + ⋯

  1. r = 4 ÷ 12 = 13. Divide consecutive terms.
  2. |r| < 1, so the series converges and S = a11-r. Check the ratio before using the formula.
  3. S = 121 - 13 = 1223 = 18. Dividing by 23 is multiplying by 32.

18

Words to know

Sequence
An ordered list of terms.
Series
The sum of the terms of a sequence.
Sigma notation
∑k=1n ak — add ak for every k from 1 to n.
Common difference
What is added each step in an arithmetic sequence.
Common ratio
What is multiplied each step in a geometric sequence.

Mistakes to watch for

Miscounting the terms. From 7 to 31 in steps of 3 there are 9 terms, not 8: 31-73 = 8 counts the gaps, and both endpoints are included so one more term exists than gaps.

Using the infinite formula when |r| ≥ 1. For r = 2 the terms grow, so there is no total. The formula still returns a number, which is why the ratio must be checked first.

Confusing the sequence with the series. The nth term is one number; the series is everything up to it added together. A question asking for "the 10th term" does not want the sum.

Mixing up the arithmetic and geometric formulas. Subtract consecutive terms to test for arithmetic, divide them to test for geometric — constant differences indicate arithmetic; constant ratios indicate geometric. If neither is constant, neither formula applies. Ratios cannot be taken through zero terms.

Dropping the a1 in the geometric sum. 1-rn1-r is the sum only when the first term is 1. For 3, 6, 12, … the whole fraction still has to be multiplied by 3.

The boy was Carl Friedrich Gauss, and he was about nine. He went on to be one of the greatest mathematicians who ever lived.

The Sigma symbol is a Greek capital sigma, for "sum". Euler introduced it in 1755 and it has not changed since.

An infinite geometric series can have a finite total: 12 + 14 + 18 + ⋯ adds to exactly 1, which is how Zeno’s paradox dissolves.

Problem 1

Evaluate ∑k=14 k.

Worked answer

10

1+2+3+4=10

Problem 2

Evaluate ∑k=13 2k.

Worked answer

12

2+4+6=12

Problem 3

Evaluate ∑k=14 (k+3).

Worked answer

22

4+5+6+7=22

Problem 4

How many terms are there from 5 to 20 counting by 5?

Worked answer

4

5, 10, 15, 20

Four terms

Problem 5

Is 3, 9, 27, 81 arithmetic or geometric? Answer with one word.

Worked answer

geometric

Each is triple the last

Problem 6

Find the sum of the whole numbers from 1 to 50.

Worked answer

1275

502(1+50)

=25 × 51 = 1275

Problem 7

Find the sum of the first 20 terms of 3, 7, 11, …

Worked answer

820

a20=3+19(4)=79

202(3+79)=10 × 82=820

Problem 8

How many terms are there from 7 to 31 counting by 3?

Worked answer

9

31-73=8 gaps

8+1=9 terms

Problem 9

Find the sum of the first 5 terms of 2, 6, 18, …

Worked answer

242

21-2431-3=2 × 121=242

Problem 10

Find the sum of the first 4 terms of 5, 10, 20, …

Worked answer

75

5+10+20+40=75

Problem 11

Find the common ratio of 48, 12, 3, …

Worked answer

1/4

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Problem 12

Find the sum of 8+4+2+1+⋯

Worked answer

16

r=12

81-12=16

Problem 13

Find the sum of 9+3+1+⋯

Worked answer

13.5

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Problem 14

Which infinite series has a finite sum?

  1. 1+12+14+⋯
  2. 1+2+4+⋯
  3. 1+1+1+⋯
  4. 2+4+8+⋯
Worked answer

1+12+14+⋯

Only the first has |r|<1

The others grow without limit

Problem 15

Write 0.4 as a fraction in lowest terms.

Worked answer

4/9

0.41-0.1=0.40.9=49

Problem 16

The first term is 6 and the sum of the first 10 terms is 195. Find the common difference.

Worked answer

3

195=5(12+9d)

39=12+9d

9d=27 so d=3

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