Sequences and Series

Find the minimum value of $x+\frac{1}{x}$.

Almost every candidate writes $2$, by the arithmetic-geometric mean inequality. Now put $x=-1$: the expression equals $-2$.

There is no minimum. The function takes every value in $(-\infty ,-2]\cup [2,\infty )$ and nothing between, so $2$ is the minimum only on the positive branch. The inequality $\frac{p+q}{2}\ge \sqrt{pq}$ is a statement about positive numbers, and the square root on the right is what makes that hypothesis unavoidable. Applied to a quantity that can be negative it does not merely give a weak answer; it gives a false one.

The same trap is set with trigonometry. Asked for the minimum of $\sin \theta +\mathrm{cosec}\theta$, the mean inequality suggests $2$, but for $\theta$ in the third or fourth quadrant the expression is at most $-2$.

1. The three progressions and the means between them

An arithmetic progression has $a_n=a+(n-1)d$ and $S_n=\frac{n}{2}\left[2a+(n-1)d\right]$. A geometric progression has $a_n=a{r}^{n-1}$ and $S_n=\frac{a\left(r^n-1\right)}{r-1}$, with $S_{\infty }=\frac{a}{1-r}$ when $|r|<1$. A harmonic progression is a sequence whose reciprocals form an AP, and it has no sum formula at all — a fact worth stating plainly, because questions exploit the expectation that one exists.

Two structural facts about an AP are used constantly. First, $S_n$ is always a quadratic in $n$ with zero constant term, and conversely any such $S_n$ describes an AP. Second, $a_n=S_n-S_{n-1}$, which recovers the terms from the sum without knowing $a$ or $d$.

The three means of two positive numbers satisfy $A\ge G\ge H$ with $G^2=AH$, so the geometric mean is the geometric mean of the other two. The relation makes any two of the three determine the pair.

One test settles which progression three numbers belong to, and it is worth carrying because questions often supply the ratio rather than the terms. For any $a$, $b$, $c$, the quotient $\frac{a-b}{b-c}$ equals $1$ when they are in AP, $\frac{a}{b}$ when they are in GP, and $\frac{a}{c}$ when they are in HP.

Each is a one-line rearrangement of the defining condition — $a-b=b-c$, $b^2=ac$, and $\frac{2}{b}=\frac{1}{a}+\frac{1}{c}$ respectively — and the pattern in the right-hand sides is easier to hold than three separate criteria.

Illustration 1

The sum of the first $n$ terms of a sequence is $S_n=3n^2+5n$. Find the $n$th term and show the sequence is an AP.

$a_n=S_n-S_{n-1}=\left(3n^2+5n\right)-\left(3n^2-6n+3+5n-5\right)=6n+2.$

The difference $a_n-a_{n-1}=6$ is constant, so it is an AP with first term $8$ and common difference $6$. The quadratic form of $S_n$ with no constant term was already a guarantee of this, and reading it off is faster than the subtraction.

Illustration 2

The arithmetic mean of two positive numbers is $5$ and their harmonic mean is $3.2$. Find the numbers.

Since $G^2=AH=16$, the geometric mean is $4$. So $a+b=10$ and $ab=16$, making $a$ and $b$ the roots of $t^2-10t+16=0$, namely $2$ and $8$.

Checking: their harmonic mean is $\frac{2\cdot 2\cdot 8}{10}=3.2$, as required.

Illustration 3

Nine arithmetic means are inserted between $2$ and $38$. Find the fifth of them.

Inserting $n$ means makes $n+2$ terms in all, so the common difference is $\frac{38-2}{n+1}=\frac{36}{10}=3.6$. The fifth mean is $2+5(3.6)=20$.

The count $n+1$, not $n$, is the step candidates get wrong: inserting nine means creates ten gaps.

2. Inserting means, and what they add up to

Inserting $n$ arithmetic means between $a$ and $b$ produces $n+2$ terms in all, so the common difference is $\frac{b-a}{n+1}$. The means themselves are symmetric about the midpoint $\frac{a+b}{2}$: pairing the first with the last, the second with the second-last, and so on, each pair sums to $a+b$. Hence

$A_1+A_2+\cdots +A_n=n\cdot \frac{a+b}{2},$

which says that $n$ inserted means contribute exactly $n$ copies of the single arithmetic mean. The corresponding statement for geometric means is multiplicative: $G_k{G}_{n+1-k}=ab$ for every $k$, so

$G_1{G}_2\cdots G_n=(ab{)}^{n/2}.$

For harmonic means the reciprocals are in AP, so it is their reciprocals that add: $\sum \frac{1}{H_k}=\frac{n(a+b)}{2ab}$. Each of these three follows from the same pairing argument, applied to sums, to products, and to reciprocals respectively, which is why they are worth deriving once rather than memorising as three separate results.

3. Telescoping and the method of differences

Most series at Advanced level are not standard progressions. They are sequences whose $n$th term can be split so that consecutive pieces cancel, and finding that split is the entire question.

The pattern to look for is a term of the form $f(r)-f(r+1)$. Any product of consecutive terms of an AP in a denominator has one, supplied by partial fractions.

Illustration 4

Sum $\sum _{r=1}^n\frac{1}{r(r+1)(r+2)}$.

Split the term as a difference of two products:

$\frac{1}{r(r+1)(r+2)}=\frac{1}{2}\left[\frac{1}{r(r+1)}-\frac{1}{(r+1)(r+2)}\right].$

Everything cancels except the first and last pieces, leaving

$\frac{1}{2}\left[\frac{1}{2}-\frac{1}{(n+1)(n+2)}\right]=\frac{1}{4}-\frac{1}{2(n+1)(n+2)}.$

Letting $n\to \infty$ gives $\frac{1}{4}$ for the infinite sum, and testing at $n=1$ gives $\frac{1}{6}$ both from the formula and from the single term.

Illustration 5

Sum $\sum _{r=1}^n\frac{1}{(3r-2)(3r+1)}$.

The denominators are consecutive terms of the AP $1,4,7,10,\dots$, so partial fractions give $\frac{1}{3}\left[\frac{1}{3r-2}-\frac{1}{3r+1}\right]$. Telescoping,

$\frac{1}{3}\left[1-\frac{1}{3n+1}\right]=\frac{n}{3n+1}.$

The factor $\frac{1}{3}$ is the common difference of the AP, and forgetting it is the usual slip.

The two illustrations above are instances of one rule. If $a_1,a_2,\dots$ is an AP with common difference $d$, then a reciprocal of $k$ consecutive terms splits as

$\frac{1}{a_r{a}_{r+1}\cdots a_{r+k-1}}=\frac{1}{(k-1)d}\left[\frac{1}{a_r\cdots a_{r+k-2}}-\frac{1}{a_{r+1}\cdots a_{r+k-1}}\right],$

so the sum telescopes with a leading factor of $\frac{1}{(k-1)d}$. For $k=2$ and $d=1$ that factor is $1$, which is why the simplest case hides it; for $k=3$ and $d=1$ it is $\frac{1}{2}$, and for $k=2$ with $d=3$ it is $\frac{1}{3}$, exactly as the two illustrations found.

Illustration 6

Sum $\sum _{r=1}^{n}r\cdot r!$.

Write $r\cdot r!=(r+1)!-r!$, which is exactly the telescoping form. The sum collapses to $(n+1)!-1$. At $n=3$ this gives $23$, matching $1+4+18$.

When the terms involve factorials, the split is almost always into consecutive factorials rather than into partial fractions.

4. Finding the $n$th term first

A series given by its first few terms must be converted into a formula before it can be summed. Take successive differences: if they are constant the term is linear in $r$, if they form an AP the term is quadratic, and if they form a GP the term contains a geometric factor.

Illustration 7

Sum the series $3+8+15+24+35+\cdots$ to $n$ terms.

The differences are $5,7,9,11$, an AP, so the $r$th term is a quadratic. Fitting, $a_r=r^2+2r=r(r+2)$, which reproduces $3,8,15,24$. Then

${\sum }_{r=1}^n\left(r^2+2r\right)=\frac{n(n+1)(2n+1)}{6}+n(n+1)=\frac{n(n+1)(2n+7)}{6}.$

Checking at $n=2$: the formula gives $\frac{2\cdot 3\cdot 11}{6}=11$, and $3+8=11$.

The three standard sums used here are $\sum r=\frac{n(n+1)}{2}$, $\sum r^2=\frac{n(n+1)(2n+1)}{6}$ and $\sum r^3={\left(\sum r\right)}^2$, the last being the identity worth remembering in its squared form rather than expanded.

5. Arithmetico-geometric series

When the $r$th term is a linear factor times a geometric factor, no rearrangement telescopes. The device is to subtract $r$ times the series from itself, which turns the linear part into a constant and leaves an ordinary GP.

For $S={\sum }_{r=1}^n\left[a+(r-1)d\right]x^{r-1}$, forming $S-xS$ gives

$S=\frac{a}{1-x}+\frac{dx\left(1-x^{n-1}\right)}{(1-x{)}^2}-\frac{\left[a+(n-1)d\right]x^n}{1-x},$

and in the infinite case with $|x|<1$ only the first two terms survive, giving $\frac{a}{1-x}+\frac{dx}{(1-x{)}^2}$.

Illustration 8

Sum $1+\frac{2}{3}+\frac{3}{9}+\frac{4}{27}+\cdots$ to infinity.

Here $a=1$, $d=1$ and $x=\frac{1}{3}$, so the sum is $\frac{1}{1-\frac{1}{3}}+\frac{\frac{1}{3}}{{\left(1-\frac{1}{3}\right)}^2}=\frac{3}{2}+\frac{\frac{1}{3}}{\frac{4}{9}}=\frac{3}{2}+\frac{3}{4}=\frac{9}{4}$.

The same answer comes from recognising the series as $\sum r{x}^{r-1}=\frac{1}{(1-x{)}^2}$, which is the derivative of the geometric series — a check worth doing, since the subtraction method is error-prone.

Illustration 9

An infinite geometric progression has sum $4$, and the sum of the cubes of its terms is $192$. Find the first term.

The cubes form a GP with first term $a^3$ and ratio $r^3$, so

$\frac{a}{1-r}=4,\frac{a^3}{1-r^3}=192.$

Substituting $a=4(1-r)$ into the second and cancelling $(1-r)$ gives $(1-r{)}^2=3\left(1+r+r^2\right)$, that is $2r^2+5r+2=0$, so $r=-\frac{1}{2}$ or $r=-2$. Convergence rules out $r=-2$, leaving $r=-\frac{1}{2}$ and $a=6$.

Checking: $\frac{6}{1.5}=4$ and $\frac{216}{1+\frac{1}{8}}=192$. Discarding the root that violates $|r|<1$ is part of the answer, not a formality.

6. The mean inequalities, used correctly

For positive reals $a_1,\dots ,a_n$,

$\frac{a_1+\cdots +a_n}{n}\ge \sqrt[n]{a_1\cdots a_n}\ge \frac{n}{\frac{1}{a_1}+\cdots +\frac{1}{a_n}},$

with equality throughout exactly when all the numbers are equal. That equality condition is what makes the inequality usable for optimisation: a bound is only a maximum or minimum if some choice attains it, and the condition tells you which choice to test.

Illustration 10

For positive $a$, $b$, $c$, show that $(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge 9$.

Apply the mean inequality to each bracket: $a+b+c\ge 3\sqrt[3]{abc}$ and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge \frac{3}{\sqrt[3]{abc}}$. Multiplying, the cube roots cancel and the product is at least $9$. Equality needs $a=b=c$ in both, which is consistent, so $9$ is attained and is genuinely the minimum.

Illustration 11

If $x$, $y$, $z$ are positive with $xyz=27$, find the least value of $x+y+z$.

The mean inequality gives $x+y+z\ge 3\sqrt[3]{27}=9$, and equality holds when $x=y=z=3$, which satisfies the constraint. So the least value is $9$.

Contrast this with the opening question, where no such attaining point exists on the whole domain. The difference is entirely in whether the hypothesis of positivity is part of the problem or has been assumed silently.

Illustration 12

If $a$, $b$, $c$ are in harmonic progression, prove that $\frac{a+b}{2a-b}+\frac{c+b}{2c-b}\ge 4$.

Being in HP means $\frac{2}{b}=\frac{1}{a}+\frac{1}{c}$, so $b=\frac{2ac}{a+c}$. Substituting turns each fraction into $\frac{a+c}{c}$ and $\frac{a+c}{a}$ respectively, whose sum is $2+\frac{a}{c}+\frac{c}{a}$. Since $\frac{a}{c}+\frac{c}{a}\ge 2$ for positive $a$ and $c$, the total is at least $4$.

Note where positivity was used: the last step fails if $a$ and $c$ have opposite signs, which is precisely the trap of the opening question appearing inside a harder problem.

One further inequality earns its place because it handles problems the mean inequality cannot: those with a sum of squares as the constraint. For any real numbers,

${\left(\sum a_i{b}_i\right)}^2\le \left(\sum a_i^2\right)\left(\sum b_i^2\right),$

with equality when the two lists are proportional. Unlike the mean inequality this needs no positivity, because everything in it is squared. A frequently used rearrangement, valid for positive $y_i$, is $\sum \frac{x_i^2}{y_i}\ge \frac{{\left(\sum x_i\right)}^2}{\sum y_i}$.

Illustration 13

If $a^2+b^2+c^2=1$, find the greatest value of $a+2b+3c$.

Take the two lists $(a,b,c)$ and $(1,2,3)$. Then

$(a+2b+3c{)}^2\le \left(a^2+b^2+c^2\right)\left(1+4+9\right)=14,$

so the expression is at most $\sqrt{14}$. Equality needs $(a,b,c)$ proportional to $(1,2,3)$, and $\left(\frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}}\right)$ satisfies the constraint and attains the bound. The mean inequality is of no use here, because $a$, $b$ and $c$ are not required to be positive.

7. Harmonic progressions and comparisons

Because reciprocals of an AP have no closed sum, every HP question is really an AP question in disguise, and the first move is always to invert. What survives from the AP is the middle-term property: $b$ is the harmonic mean of $a$ and $c$, so $b=\frac{2ac}{a+c}$, and this sits below the geometric mean $\sqrt{ac}$ and below the arithmetic mean $\frac{a+c}{2}$.

The chain $A\ge G\ge H$ collapses only when the two numbers are equal, which gives a quick way to answer a family of short questions: if three numbers are in arithmetic, geometric and harmonic progression all at once, they must be identical, since $A=G$ already forces it.

For more than two numbers the relation $G^2=AH$ fails — it is genuinely a two-number identity — while the ordering $A\ge G\ge H$ continues to hold for any count, with equality again only when all the numbers agree.

Illustration 14

If $a$, $b$, $c$ are in AP and $a$, $b^{\prime }$, $c$ are in GP and $a$, $b^{\prime \prime }$, $c$ are in HP with $a$ and $c$ positive and unequal, order $b$, $b^{\prime }$ and $b^{\prime \prime }$.

They are the arithmetic, geometric and harmonic means of the same pair, so $b>b^{\prime }>b^{\prime \prime }$, strictly because $a\ne c$. The relation $b^{\prime 2}=b{b}^{\prime \prime }$ also holds, so the three means are themselves in geometric progression.

Illustration 15

Prove that in any AP with positive terms, $\frac{1}{\sqrt{a_1}+\sqrt{a_2}}+\frac{1}{\sqrt{a_2}+\sqrt{a_3}}+\cdots +\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_n}}=\frac{n-1}{\sqrt{a_1}+\sqrt{a_n}}$.

Rationalise each term: $\frac{1}{\sqrt{a_k}+\sqrt{a_{k+1}}}=\frac{\sqrt{a_{k+1}}-\sqrt{a_k}}{d}$, where $d$ is the common difference. The sum telescopes to $\frac{\sqrt{a_n}-\sqrt{a_1}}{d}$, and rationalising once more, using $a_n-a_1=(n-1)d$, gives the stated form.

This combines the two main devices of the chapter: rationalising to create a difference, then telescoping it.

Summary

The mean inequalities hold for positive numbers only, and applying them without that hypothesis produces answers that are wrong rather than merely weak: $x+\frac{1}{x}$ has no minimum on the reals. Equality in $A\ge G\ge H$ requires all the numbers to be equal, and checking that the equality case is attainable is what turns a bound into a maximum or minimum.

An AP has $S_n$ quadratic in $n$ with zero constant term, and $a_n=S_n-S_{n-1}$ recovers the terms. Inserting $n$ means creates $n+1$ gaps. For two positive numbers $G^2=AH$, so any two of the three means determine the pair. A harmonic progression has no sum formula, so every HP question begins by inverting to an AP.

Series that are not progressions are summed by telescoping. Look for a term of the form $f(r)-f(r+1)$: partial fractions supply it when consecutive terms of an AP sit in a denominator, consecutive factorials supply it when factorials appear, and rationalising supplies it when square roots do. If a series is given only by its first few terms, take successive differences to identify the form of the $r$th term before summing.

Inserting $n$ means contributes $n$ copies of the single mean, additively for arithmetic means, multiplicatively for geometric ones, and reciprocally for harmonic ones. When the constraint is a sum of squares rather than a product, the Cauchy inequality applies where the mean inequality cannot, and it needs no positivity because every quantity in it is squared.

An arithmetico-geometric series is summed by subtracting $x$ times the series from itself, and the infinite case reduces to $\frac{a}{1-x}+\frac{dx}{(1-x{)}^2}$. In any infinite geometric problem the condition $|x|<1$ is part of the answer, and roots violating it must be discarded explicitly.