Sequences and Series

Written and reviewed by Devendra Dubey, Mathematics faculty at UniDrill (M.Sc. Mathematics, Osmania University · 10 years teaching Mathematics). Based on NCERT Mathematics Class XI, Chapter 8.

How many ancestors do you have?

You have 2 parents, 4 grandparents, 8 great-grandparents, and each generation back doubles the count. Ten generations back, that single generation has $2^{10} = 1024$ people. Add up every generation from your parents to that one and you get 2046 people. A list that grows by the same multiplier each step is a geometric progression, and the shortcut for adding it up is the heart of this chapter.

Sequences: an ordered list with a rule

A sequence is a list of numbers in a definite order: $a_1, a_2, a_3, \ldots, a_n, \ldots$. The subscript is the position and $a_n$ is the general term (or $n$-th term). Order matters: 1, 2, 3 and 3, 2, 1 are different sequences. A sequence is really a function whose domain is the natural numbers, which is why you will see it again in the chapter on relations and functions.

A sequence with a finite number of terms is finite, otherwise it is infinite. The rule that generates the terms can take different forms.

A recurrence relation defines each term from the earlier ones, so you have to build the sequence step by step. Using the Fibonacci rule, the 7th term is $8 + 5 = 13$.

Series and sigma notation

A series is the indicated sum of the terms of a sequence: $a_1 + a_2 + \cdots + a_n + \cdots$. Sigma notation writes it compactly:

$$\sum_{k=1}^{n} a_k = a_1 + a_2 + \cdots + a_n$$

The word "series" names the expression, not the number it equals. So $1 + 3 + 5 + 7$ is a four-term series, and its sum is 16.

Worked example: series versus sum. Write the series $\displaystyle\sum_{k=1}^{4}(2k+1)$ in expanded form and find its sum.

Common wrong step: answering "24" when asked for the series. The series is the indicated sum $3+5+7+9$; 24 is its sum.

Geometric progression: constant multiplier

A sequence $a_1, a_2, \ldots$ with all terms non-zero is a geometric progression (GP) if the ratio of each term to the one before is the same constant $r$, called the common ratio: $\dfrac{a_{k+1}}{a_k} = r$. With first term $a$, the GP is $a,\ ar,\ ar^2,\ ar^3,\ldots$

The general term. The first term is $a$ (that is $ar^0$), the second is $ar^1$, and so the $n$-th term has the power $n-1$:

$$a_n = a\,r^{\,n-1}$$

Worked example: which term? Which term of the GP 2, 8, 32, … is 131072?

Common wrong step: writing $2\cdot4^{n} = 131072$ and getting $n = 8$. The exponent is $n-1$, not $n$.

Worked example: recover the GP from two terms. In a GP the third term is 12 and the sixth term is 96. Find the first term, the common ratio and the 8th term.

Common wrong step: concluding $r = 96 \div 12 = 8$. The two terms are three steps apart, so their ratio is $r^3$, not $r$.

Sum of a GP: the subtract-and-cancel trick

Write the sum and multiply it by $r$:

$$S_n = a + ar + ar^2 + \cdots + ar^{n-1}$$ $$rS_n = \;\;\;\;ar + ar^2 + \cdots + ar^{n-1} + ar^{n}$$

Subtract. Everything cancels except the first and last terms, so $(r-1)S_n = a(r^n - 1)$, and

$$S_n = \dfrac{a\,(r^{n}-1)}{r-1} = \dfrac{a\,(1-r^{n})}{1-r} \quad (r \ne 1), \qquad S_n = na \quad (r=1)$$

The two fractions are identical. Pick the one where the top and bottom are positive: use $r^n-1$ over $r-1$ when $r>1$, and $1-r^n$ over $1-r$ when $r<1$. When $r=1$ every term is $a$, so the formula would divide by zero and you must use $na$ instead. For example, $1+\tfrac12+\tfrac14+\cdots$ with 6 terms has $r=\tfrac12$: $S_6 = \dfrac{1-\left(\tfrac12\right)^6}{1-\tfrac12} = \dfrac{63}{32}$.

Worked example: the ancestors. Find the total number of ancestors over 10 generations.

Common wrong step: using $r=1$ in the denominator by mistake, or computing only the last generation, $2^{10} = 1024$. The question asks for the total.

Pause and predict: a GP has first term 3 and common ratio 1. What is the sum of its first 10 terms, and why can you not use $\dfrac{a(r^n-1)}{r-1}$?

Check your answer: every term is 3, so the sum is $10 \times 3 = 30$. The formula gives $0/0$ when $r=1$, so the separate result $S_n = na$ is required.

Beyond the textbook. For the exam: the sum formula is applied to doubling and halving problems. In reality: money that grows at a fixed rate each year follows a GP. If ₹10000 grows at 10% a year, the amounts at the end of each year are $10000(1.1),\ 10000(1.1)^2,\ 10000(1.1)^3, \ldots$, which is a GP with ratio 1.1, so after 3 years you have ₹13310. Savings plans that add the same deposit every period are worked out by summing a GP in exactly this way.

Geometric mean: the number that fits in the middle

Three positive numbers $a, G, b$ are in GP when $\dfrac{G}{a} = \dfrac{b}{G}$, so $G^2 = ab$. The geometric mean of two positive numbers is therefore

$$G = \sqrt{ab}$$

The GM of 2 and 8 is $\sqrt{16} = 4$, and 2, 4, 8 is indeed a GP (ratio 2). The GM is defined here for positive numbers, and it is the positive root.

Inserting $n$ geometric means. To put $n$ numbers $G_1, G_2, \ldots, G_n$ between $a$ and $b$ so that $a, G_1, \ldots, G_n, b$ is a GP, count the terms: there are $n+2$ in all, and $b$ is the last one, term number $n+2$, so $b = ar^{\,n+1}$. Therefore

$$r = \left(\dfrac{b}{a}\right)^{\frac{1}{n+1}}, \qquad G_k = a\,r^{k}$$

Worked example: insert three geometric means. Insert three numbers between 1 and 256 so that all five form a GP with positive common ratio.

Common wrong step: writing $r^{3} = 256$ because three numbers are inserted. The ratio is applied once for every gap, and there are $n+1 = 4$ gaps.

Arithmetic mean and the AM–GM inequality

The arithmetic mean of $a$ and $b$ is $A = \dfrac{a+b}{2}$. For positive numbers it is never smaller than the geometric mean. The proof is one line:

$$A - G = \dfrac{a+b}{2} - \sqrt{ab} = \dfrac{\left(\sqrt a - \sqrt b\right)^{2}}{2} \ge 0$$

A square is never negative, so $A \ge G$, with equality exactly when $\sqrt a = \sqrt b$, that is $a = b$.

Worked example: find two numbers from their means. The AM of two positive numbers is 10 and their GM is 8. Find the numbers.

Common wrong step: using $a + b = 10$, which is the AM itself. The AM is half the sum. Check: $\dfrac{4+16}{2} = 10$ and $\sqrt{4\times16} = 8$.

Pause and predict: for positive $x$, what is the smallest possible value of $x + \dfrac{16}{x}$?

Check your answer: apply AM $\ge$ GM to $x$ and $\dfrac{16}{x}$: $\dfrac{x + 16/x}{2} \ge \sqrt{16} = 4$, so $x + \dfrac{16}{x} \ge 8$. Equality holds when $x = \dfrac{16}{x}$, i.e. $x=4$.

Beyond the textbook. For the exam: $A \ge G$ is proved for two positive numbers. In reality: it explains why, among all rectangles with the same area, the square has the smallest perimeter. For area 36, the sides $a$ and $b$ satisfy $a+b \ge 2\sqrt{ab} = 12$, so the perimeter is at least 24, reached only when $a = b = 6$. Designers use the same idea whenever they want to enclose a fixed area with the least fencing.

Beyond the textbook. For the exam: the Fibonacci sequence is a recurrence example. In reality: it comes from a rabbit-breeding puzzle that Leonardo of Pisa (Fibonacci) set in his book of 1202. The ratio of consecutive terms settles near 1.618: $\tfrac{8}{5} = 1.6,\ \tfrac{13}{8} = 1.625,\ \tfrac{21}{13} \approx 1.615$. The number of spirals on sunflower heads and pine cones is very often a pair of consecutive Fibonacci numbers.

Exam traps

Formula sheet

60-second revision

Where this chapter connects

• Previous: Binomial Theorem gives a finite series whose terms are binomial coefficients times powers.
• Next: Straight Lines begins coordinate geometry; the idea of constant change reappears as slope.
• Same strand: Relations and Functions treats a sequence as a function on the natural numbers.
• Later: Limits and Derivatives asks what happens to $a_n$ as $n$ grows without bound.