Real sequences

What Nour says

1. 1. Bounded sequences and monotone sequences

   Today, the whole chapter: "Real sequences", Year 4, Mathematics section.

   We will build each idea step by step, with a picture and a worked example every time.

   By the end, you will handle recurrent sequences, limits, and bac-style sequence problems with confidence.

   Let's start with monotone sequences and bounded sequences.

   A sequence is just a list of numbers, one for each whole number n.

   We say a sequence is increasing when each term is bigger than or equal to the one before.

   We say it is decreasing when each term is smaller than or equal to the one before.

   Together, increasing and decreasing sequences are called monotone sequences.

   Now let's talk about bounded sequences, suite bornée in the manual's words.

   That is, a sequence is bounded when all its terms stay between two fixed numbers.

   M is called an upper bound, and m is called a lower bound.

   Picture the terms as points on a number line, trapped inside a horizontal band.

   Here is a very important special case: increasing and bounded above.

   This combination has its own name: "croissante et majorée", that is, increasing and bounded above.

   Here is why this case matters so much in this chapter.

   An increasing sequence keeps climbing, but if it is bounded above, it cannot climb forever.

   So it gets closer and closer to some limit value, without ever crossing the ceiling M.

   This idea is the key theorem of this section: every increasing, bounded above sequence converges.

   The same holds the other way: a decreasing sequence bounded below also converges.

   Let's now work through a full example, the kind you meet in bac exercises.

   Consider the sequence defined for n greater than or equal to one by this formula.

   First, let's show this sequence is increasing, by comparing u(n+1) and u(n).

   We compute the difference u(n+1) minus u(n) and simplify it into one fraction.

   After putting both fractions on the same denominator, the numerator simplifies nicely.

   This fraction is always positive, because n is a positive whole number.

   So u(n+1) minus u(n) is positive, which means the sequence is increasing.

   Now let's check if this sequence is bounded above.

   We rewrite u(n) by splitting the fraction to see its behaviour clearly.

   Since one over n plus one is always positive, u(n) stays strictly below two.

   So two is an upper bound, and the sequence is bounded above.

   We now have both conditions: increasing, and bounded above by two.

   By our theorem, this sequence must converge to some limit.

   In fact here we can even guess the limit directly from the rewritten formula.

   As n grows large, one over n plus one gets closer and closer to 0.

   So the limit of this sequence is exactly two.

   Let's gather the key result of this section in a few words.

   A bounded sequence has all its terms trapped between a lower and an upper bound.

   A monotone sequence is always increasing, or always decreasing, never both.

   And the big theorem: increasing plus bounded above always gives a convergent sequence.

   Why must an increasing sequence that is bounded above converge?

2. 2. Recurrent sequences

   Let's move to recurrent sequences.

   This is one of the manual's key terms: suites récurrentes, that is, sequences built from their own previous term.

   We fix a first term u zero, then a rule f tells us how to get the next term.

   This is different from a sequence given by a direct formula in n.

   Here we must climb term by term, we cannot jump straight to u(100).

   Here is why this matters: to study such a sequence, we often study the function f first.

   If f is increasing, the sequence keeps the same order as it moves along.

   A useful trick is to look for a fixed point, where f of l equals l.

   If the sequence converges, its limit must be a solution of this equation.

   Let's work through a full bac-style example now.

   Take u zero equal to zero, and u(n+1) equal to one half times u(n), plus one.

   First, let's show by induction that u(n) stays below two for every n.

   The base case is clear: u zero equals zero, which is less than two.

   Now assume u(n) is less than two, and let's check u(n+1).

   Since u(n) is less than two, one half of u(n) is less than one.

   Adding one gives u(n+1) less than two, so the property passes to the next term.

   By induction, u(n) stays below two for every whole number n.

   Next, let's show the sequence is increasing, using this same bound.

   We compute u(n+1) minus u(n), which equals one minus one half of u(n).

   Since u(n) is less than two, one half of u(n) is less than one.

   So one minus one half of u(n) is positive, and the sequence is increasing.

   We now have increasing, and bounded above by two.

   By the theorem from our last section, this sequence converges to some limit l.

   To find l, we use the fixed point idea: l equals one half of l, plus one.

   Solving this equation gives l equal to 2.

   So this recurrent sequence increases and converges to two.

   Let's gather the key result of this section.

   A recurrent sequence is built step by step, each term from the one before it.

   To study it, we check monotony and boundedness, often by induction.

   If it converges, its limit solves the fixed-point equation f of l equals l.

   In this example, what equation must the limit l satisfy?

3. 3. Limits and comparison theorems

   Let's now move to limits and comparison theorems.

   These are called comparison theorems (théorèmes de comparaison), that is, rules to find a limit by comparing sequences.

   Sometimes we cannot compute a limit directly, but we can trap the sequence between two others.

   The first rule: if u(n) is less than or equal to w(n), and w(n) tends to minus infinity, so does u(n).

   Symmetrically, if v(n) is less than or equal to u(n), and v(n) tends to plus infinity, so does u(n).

   The strongest rule is the squeeze theorem, when u(n) is trapped between v(n) and w(n).

   Here is why this works: if both fences meet at the same value l, u(n) has no choice.

   It gets squeezed tighter and tighter, so it must also reach that same limit l.

   Picture two curves closing in on each other, with our sequence trapped between them.

   Let's solve a full bac-style example using this squeeze theorem.

   Consider u(n) equal to sine of n, divided by n, plus one, for n at least one.

   We know that sine of n always stays between minus one and one.

   Dividing everything by n, which is positive, keeps the same order of the inequalities.

   This gives minus one over n, less than or equal to sine of n over n, less than or equal to one over n.

   Adding one everywhere gives a frame for u(n) itself.

   Now both fences, one minus one over n and one plus one over n, tend to one.

   By the squeeze theorem, u(n) is forced to tend to one as well.

   We never needed the exact value of sine of n to find this limit.

   This shows the real strength of comparison theorems, in bac problems with sine or cosine.

   Let's gather the key result of this section.

   Comparison theorems let us find a limit without computing the sequence directly.

   The squeeze theorem traps u(n) between two sequences that both tend to the same limit.

   This is especially powerful whenever sine, cosine, or any bounded term appears in u(n).

   In the squeeze theorem, what must be true about the two bounding sequences?

4. 4. Adjacent sequences

   Let's move to adjacent sequences.

   In French, these are called suites adjacentes, that is, two sequences that come closer and closer together.

   We need two sequences, u(n) and v(n), with three conditions together.

   First, u(n) is increasing, and second, v(n) is decreasing.

   Third, the difference v(n) minus u(n) tends to zero.

   When these three conditions hold, both sequences converge, and they share the same limit.

   Here is why this makes sense: u(n) climbs up, v(n) comes down, and the gap shrinks to zero.

   So they must be closing in on exactly the same point on the number line.

   Picture u(n) rising from below, and v(n) falling from above, meeting near the same limit.

   Let's now work through a complete bac-style example.

   Define u(n) equal to one plus one over one factorial, up to one over n factorial.

   And define v(n) equal to u(n) plus one over n times n factorial.

   Let's check our first condition: is u(n) increasing.

   u(n+1) minus u(n) equals one over n plus one factorial, which is always positive.

   So u(n) is increasing, our first condition holds.

   Now let's check the second condition: is v(n) decreasing.

   After simplifying v(n+1) minus v(n), the sign turns out to be negative.

   So v(n) is decreasing, our second condition holds too.

   Finally, let's check the third condition: the gap v(n) minus u(n).

   By definition, v(n) minus u(n) equals one over n times n factorial.

   As n grows large, n factorial grows huge, so this fraction tends to zero.

   All three conditions hold, so u(n) and v(n) are adjacent sequences.

   They converge to the same limit l, which happens to be the number e.

   Let's gather the key result of this section.

   Adjacent sequences need one increasing, one decreasing, and a gap tending to zero.

   When all three hold, both sequences converge to exactly the same limit.

   What must happen to the gap v(n) minus u(n) for two sequences to be adjacent?

5. 5. Sequences defined by an integral

   Let's finish with sequences defined by an integral.

   The manual calls these sequences defined by an integral (suites définies par une intégrale), that is, each term u(n) is an area under a curve.

   A typical form is u(n) equal to the integral from zero to one of f sub n of x, dx.

   Here n sits inside the function, so each term is the area under a slightly different curve.

   We cannot always compute the exact value, but we can still study the sequence's behaviour.

   Here is why comparison helps so much here: inequalities on functions give inequalities on the areas.

   If one curve always stays under another on the interval, its area stays smaller too.

   This lets us trap u(n) between simpler sequences, and reuse our comparison theorems.

   Let's work through a full bac-style example now.

   Define u(n) as the integral from zero to one of x to the power n, dx.

   First, let's show that u(n) is always between zero and one.

   On the interval from zero to one, x to the power n stays between zero and one.

   Integrating this inequality over the interval gives zero less than or equal to u(n), less than or equal to one.

   So u(n) is a bounded sequence, exactly the term we studied earlier.

   Next, let's show u(n) is decreasing, by comparing x to the power n and n plus one.

   On zero to one, x to the power n plus one is always less than or equal to x to the power n.

   Integrating again gives u(n+1) less than or equal to u(n), so the sequence decreases.

   So u(n) is decreasing, and bounded below by zero.

   By our first theorem, decreasing and bounded below, this sequence converges.

   We can also compute u(n) directly, as one over n plus one.

   As n grows large, this fraction tends to zero.

   Let's gather the key result of this whole chapter.

   Sequences defined by an integral behave like ordinary sequences, once we compare the functions inside them.

   Monotony and bounds often come from simple inequalities on the function, before integrating.

   Why does the inequality on x to the power n give us a bound on u(n)?

   That completes the whole chapter on real sequences, from bounds to integrals.

   You now have every tool the bac expects: monotony, comparison, adjacent sequences, and recurrence.

   Practice the two exercises above, and these ideas will feel completely natural.