Part 1 · Chapter 04

Sequences & Series

Patterns that march in step, multiply, or converge — and the means and sums that tie them together

Fundamentals of Mathematics Prof. Mithun Mondal Reading time ≈ 38 min
i What you'll learn
  • The \(n\)th term and sum of an arithmetic and a geometric progression.
  • When an infinite geometric series converges, and what it sums to.
  • Harmonic progressions, and the three means — AM, GM, HM — with the inequality that orders them.
  • To sum an arithmetico-geometric series using the shift-and-subtract trick.
  • The power sums \(\sum n,\ \sum n^{2},\ \sum n^{3}\) and how to combine them.
  • The method of differences (telescoping) for sums that collapse.
Section 4-1

Sequences & Series: The Setup

A sequence is an ordered list of numbers following a rule — formally, a function whose domain is the natural numbers. We write its terms \(a_1,a_2,a_3,\dots\) and call \(a_n\) the general term. A series is what you get when you add the terms of a sequence:

Sequence vs series
\[ \underbrace{a_1,\,a_2,\,a_3,\dots}_{\text{sequence}}\qquad\longrightarrow\qquad \underbrace{S_n=a_1+a_2+\dots+a_n=\sum_{k=1}^{n}a_k}_{\text{series (partial sum)}} \]
A key link: the \(n\)th term is recovered from the sums by \(a_n=S_n-S_{n-1}\).
Three progressions rule this chapter. In an arithmetic progression each term adds a fixed amount; in a geometric one each term multiplies by a fixed ratio; in a harmonic one the reciprocals form an arithmetic progression. Master these three and their means, and most series problems fall open.
Section 4-2

Arithmetic Progressions (AP)

An arithmetic progression adds a constant common difference \(d\) at each step. The terms sit at equal spacings on the number line:

aa+da+2da+3da+4d +d+d+d+d
An AP: a constant step d between consecutive terms
➕
nth term and sum of an AP
\(a_n=a+(n-1)d\qquad S_n=\dfrac{n}{2}\big[\,2a+(n-1)d\,\big]=\dfrac{n}{2}(a+l)\)

Here \(l=a_n\) is the last term. The second form — half the count times (first + last) — is the famous trick young Gauss used to add \(1+2+\dots+100\) in seconds.

The defining property. Three numbers \(a,b,c\) are in AP iff \(2b=a+c\) — each term is the average of its neighbours. A neat solving tip: choose symmetric terms like \(a-d,\,a,\,a+d\) so the unknowns cancel.
Section 4-3

Arithmetic Mean & Inserting Means

The arithmetic mean of \(a\) and \(b\) is \(A=\dfrac{a+b}{2}\) — the single number that sits between them in AP. To insert \(n\) arithmetic means between \(a\) and \(b\), build an AP with \(n+2\) terms:

Inserting n arithmetic means
\[ d=\frac{b-a}{n+1},\qquad \sum_{k=1}^{n}A_k=n\cdot\frac{a+b}{2} \]
The means form an AP with that common difference; their sum is n times the single AM of the endpoints.
Section 4-4

Geometric Progressions (GP)

A geometric progression multiplies by a constant common ratio \(r\) at each step. Growth (or decay) compounds:

aarar²ar³ar⁴ each bar = r × previous
A GP with r > 1: terms scale by the ratio r
✖️
nth term and sum of a GP
\(a_n=ar^{\,n-1}\qquad S_n=a\,\dfrac{r^{n}-1}{r-1}=a\,\dfrac{1-r^{n}}{1-r}\ \ (r\neq1)\)

Use the first form when \(r>1\), the second when \(r<1\) — they are identical, just arranged to keep things positive.

The defining property. Three non-zero numbers \(a,b,c\) are in GP iff \(b^{2}=ac\) — each term is the geometric mean of its neighbours. Symmetric choice: \(\dfrac{a}{r},\,a,\,ar\).
Section 4-5

Infinite Geometric Series

If the ratio is small enough that terms shrink toward zero, an endless sum can still land on a finite value. The partial sums creep up to a limit:

aarar² sum = a/(1−r)
For |r| < 1 the shrinking pieces fill a finite length
♾️
Sum to infinity
If \(|r|<1\), then \(\ S_\infty=a+ar+ar^{2}+\cdots=\dfrac{a}{1-r}\)

If \(|r|\ge1\) the terms do not shrink and the series diverges — there is no finite sum. This single formula converts recurring decimals to fractions: \(0.\overline{3}=\tfrac{3/10}{1-1/10}=\tfrac13\).

Section 4-6

Geometric Mean & Inserting Means

The geometric mean of two positive numbers is \(G=\sqrt{ab}\). To insert \(n\) geometric means between \(a\) and \(b\), build a GP of \(n+2\) terms:

Inserting n geometric means
\[ r=\left(\frac{b}{a}\right)^{\frac{1}{n+1}},\qquad \prod_{k=1}^{n}G_k=(ab)^{n/2}=(\sqrt{ab}\,)^{\,n} \]
The product of the n inserted GMs is the single GM of the endpoints, raised to the nth power.
Section 4-7

Harmonic Progressions (HP)

A harmonic progression is a sequence whose reciprocals form an AP. There is no tidy closed form for its sum; instead, you flip to the AP, work there, and flip back.

🔁
Harmonic progression & mean
\(a,b,c\) are in HP \(\iff\dfrac1a,\dfrac1b,\dfrac1c\) are in AP \(\iff b=\dfrac{2ac}{a+c}\)

That value \(\dfrac{2ac}{a+c}\) is the harmonic mean of \(a\) and \(c\). The \(n\)th term of an HP is just the reciprocal of the corresponding AP's \(n\)th term.

Section 4-8

The AM–GM–HM Inequality

For two positive numbers the three means are always ordered the same way, and the semicircle below proves it at a glance: build a semicircle on a diameter of length \(a+b\). The radius is the AM, a half-chord is the GM, and a projected segment is the HM.

AB PO Q AM GM HM AP = a PB = b
Radius = AM, half-chord = GM, projection = HM — so AM ≥ GM ≥ HM
⚖️
The mean inequality
For positive reals: \(\ \text{AM}\ \ge\ \text{GM}\ \ge\ \text{HM}\), with equality iff all the numbers are equal.

For two numbers the means also satisfy \(\text{GM}^{2}=\text{AM}\cdot\text{HM}\). The general AM–GM states \(\dfrac{a_1+\dots+a_n}{n}\ge\sqrt[n]{a_1\cdots a_n}\) — one of the most useful inequalities in all of optimisation.

Section 4-9

Arithmetico-Geometric Series

Multiply an AP term-by-term with a GP and you get an arithmetico-geometric progression (AGP), with general term \([a+(n-1)d]\,r^{\,n-1}\). To sum it, use the shift-and-subtract trick: write \(S\), multiply by \(r\) to get \(rS\), line them up shifted by one place, and subtract.

🔗
Sum to infinity of an AGP
If \(|r|<1\): \(\ S_\infty=\dfrac{a}{1-r}+\dfrac{d\,r}{(1-r)^{2}}\)

The first piece is the GP-style term; the second captures the steadily growing AP factor. For a finite sum, run the same \(S-rS\) subtraction and keep the boundary terms.

Section 4-10

Sums of Powers: Σn, Σn², Σn³

Three power sums appear constantly; memorise them and most "find the sum of the series" problems reduce to plugging in.

Table 4-1 · The standard power sums
SeriesClosed form
\(\displaystyle\sum_{k=1}^{n}k\)\(\dfrac{n(n+1)}{2}\)
\(\displaystyle\sum_{k=1}^{n}k^{2}\)\(\dfrac{n(n+1)(2n+1)}{6}\)
\(\displaystyle\sum_{k=1}^{n}k^{3}\)\(\left[\dfrac{n(n+1)}{2}\right]^{2}=\Big(\sum k\Big)^{2}\)
A pretty fact. The sum of the first \(n\) cubes is the square of the sum of the first \(n\) naturals — so \(1^3+2^3+\dots+n^3=(1+2+\dots+n)^2\). To sum a polynomial series, split it into these standard sums by linearity.
Section 4-11

Method of Differences (Telescoping)

If each term can be written as a difference of consecutive values of some function — \(a_k=f(k)-f(k+1)\) — almost everything cancels and only the boundary survives. This is the method of differences, or telescoping.

A telescoping classic
\[ \sum_{k=1}^{n}\frac{1}{k(k+1)}=\sum_{k=1}^{n}\left(\frac{1}{k}-\frac{1}{k+1}\right)=1-\frac{1}{n+1}=\frac{n}{n+1} \]
Every middle term is added then subtracted; only the first and last endpoints remain.
! Spotting a telescoper

When a term is a product of an AP-style chain in the denominator (like \(\tfrac{1}{k(k+1)(k+2)}\)), try partial fractions to split it into a difference. If the pieces line up shifted by one or two places, the sum collapses.

Worked Examples

Putting It to Work

1 An AP from two terms

Problem. The 7th term of an AP is 34 and the 13th is 64. Find the first term, the common difference, and the sum of the first 20 terms.

Solution. \(a+6d=34\) and \(a+12d=64\). Subtracting: \(6d=30\Rightarrow d=5\), so \(a=34-30=4\). Then

Sum of 20 terms
\[ S_{20}=\frac{20}{2}\big[2(4)+19(5)\big]=10(8+95)=1030 \]
2 An infinite GP and a recurring decimal

Problem. Find the sum of \(9+3+1+\tfrac13+\cdots\), and express \(0.\overline{45}\) as a fraction.

Solution. The series has \(a=9,\ r=\tfrac13\), so \(S_\infty=\dfrac{9}{1-1/3}=\dfrac{9}{2/3}=\dfrac{27}{2}\). For the decimal, \(0.\overline{45}=\dfrac{45/100}{1-1/100}=\dfrac{45}{99}=\dfrac{5}{11}\).

3 AM–GM in action

Problem. For positive \(x\), find the minimum value of \(x+\dfrac{4}{x}\).

Solution. By AM–GM on the two positive terms, \(\dfrac{x+4/x}{2}\ge\sqrt{x\cdot\tfrac4x}=2\), so \(x+\dfrac4x\ge4\). Equality holds when \(x=\tfrac4x\), i.e. \(x=2\). The minimum is 4.

4 Summing an AGP

Problem. Find \(S=1+\dfrac{2}{3}+\dfrac{3}{3^{2}}+\dfrac{4}{3^{3}}+\cdots\) to infinity.

Solution. This is an AGP with \(a=1,\ d=1,\ r=\tfrac13\). Using \(S_\infty=\dfrac{a}{1-r}+\dfrac{dr}{(1-r)^2}\):

Working
\[ S=\frac{1}{1-\tfrac13}+\frac{1\cdot\tfrac13}{(1-\tfrac13)^{2}}=\frac{1}{\tfrac23}+\frac{\tfrac13}{\tfrac49}=\frac32+\frac34=\frac94 \]
5 A polynomial series via power sums

Problem. Find \(\displaystyle\sum_{k=1}^{n}k(k+2)\).

Solution. Expand and split: \(\sum k(k+2)=\sum k^{2}+2\sum k\). Substitute the standard sums:

Working
\[ \frac{n(n+1)(2n+1)}{6}+2\cdot\frac{n(n+1)}{2}=\frac{n(n+1)(2n+1)}{6}+n(n+1)=\frac{n(n+1)(2n+7)}{6} \]
6 A telescoping sum

Problem. Evaluate \(\displaystyle\sum_{k=1}^{n}\frac{1}{(2k-1)(2k+1)}\).

Solution. Partial fractions: \(\dfrac{1}{(2k-1)(2k+1)}=\dfrac12\left(\dfrac{1}{2k-1}-\dfrac{1}{2k+1}\right)\). The sum telescopes:

Working
\[ \frac12\left(1-\frac{1}{2n+1}\right)=\frac{n}{2n+1} \]
Review

Chapter Summary

✓ AP

\(a_n=a+(n-1)d\), \(S_n=\tfrac n2(a+l)\). Three terms in AP iff \(2b=a+c\).

✓ GP

\(a_n=ar^{n-1}\), \(S_n=a\tfrac{r^n-1}{r-1}\); for \(|r|<1\), \(S_\infty=\tfrac{a}{1-r}\). GP iff \(b^2=ac\).

✓ HP & means

Reciprocals of an AP. \(\text{AM}=\tfrac{a+b}{2}\), \(\text{GM}=\sqrt{ab}\), \(\text{HM}=\tfrac{2ab}{a+b}\).

✓ Mean inequality

\(\text{AM}\ge\text{GM}\ge\text{HM}\), equality iff all equal; and \(\text{GM}^2=\text{AM}\cdot\text{HM}\).

✓ AGP & power sums

Shift-and-subtract for AGPs; \(\sum k,\ \sum k^2,\ \sum k^3\) by standard formulas (and \(\sum k^3=(\sum k)^2\)).

✓ Telescoping

Write \(a_k=f(k)-f(k+1)\) (often via partial fractions); the middle cancels, leaving the endpoints.

Practice

Problems

Identify the progression first, then reach for its formula. Difficulty rises down the list.

  1. How many terms of the AP \(3,7,11,\dots\) add up to 210?
  2. The sum of the first \(n\) terms of a series is \(S_n=3n^{2}+2n\). Find the \(n\)th term and show the series is an AP.
  3. Insert three arithmetic means between 7 and 23.
  4. Find the sum to infinity of \(\dfrac{2}{5}+\dfrac{4}{25}+\dfrac{6}{125}+\cdots\).
  5. The 3rd term of a GP is 24 and the 6th is 192. Find the GP and the sum of its first 8 terms.
  6. If \(a,b,c\) are in HP, prove that \(\dfrac{a}{b+c},\ \dfrac{b}{c+a},\ \dfrac{c}{a+b}\) are in HP.
  7. If the AM of two positive numbers is 10 and their GM is 8, find the numbers.
  8. Prove that for positive \(a,b,c\), \((a+b+c)\left(\dfrac1a+\dfrac1b+\dfrac1c\right)\ge9\).
  9. Sum the series \(1\cdot2+2\cdot3+3\cdot4+\cdots\) to \(n\) terms.
  10. Find \(\displaystyle\sum_{k=1}^{n}\dfrac{1}{k(k+1)(k+2)}\) using partial fractions.
  11. Sum the AGP \(1+3x+5x^{2}+7x^{3}+\cdots\) to \(n\) terms \((x\neq1)\).
  12. If \(a,b,c\) are in AP, \(b,c,d\) in GP, and \(c,d,e\) in HP, show that \(a,c,e\) are in GP.
Tip: if a series mixes an arithmetic factor with a geometric one, it is an AGP — use shift-and-subtract. If the denominator is a product of consecutive terms, suspect telescoping. Naming the structure is 80% of the solution.