Math

Boundedness of convergent sequence.

Step-by-step mathematics solution: Boundedness of convergent sequence.

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1. What is being asked?

We have a convergent sequence ((a_n)_{n\in\mathbb N}) of real numbers.
We must show that every convergent sequence is bounded; i.e. there is a real number (M>0) such that

[ |a_n|\le M\qquad\text{for every }n\in\mathbb N . ]

The student wonders whether we can simply take

[ M=\max{|a_1|,|a_2|,\dots} ]

(the “maximum of all the absolute values”) and, if a maximum does not exist, take a number larger than that “supposed” maximum.
We have to explain why this idea does not work as a proof and give a correct proof.

2. Detailed proof that a convergent sequence is bounded

Let ((a_n)) be a sequence of real numbers that converges to a limit (L\in\mathbb R).
By definition of convergence:

For every (\varepsilon>0) there exists an integer (N) such that
[ n\ge N \;\Longrightarrow\; |a_n-L|<\varepsilon . ]

We will use (\varepsilon=1).

Step 1 – control the “tail’’ of the sequence

Choose (\varepsilon=1).
Then there is an integer (N) with

[ n\ge N \;\Longrightarrow\; |a_n-L|<1 . ]

For those indices we can bound ( a_n ) as follows:

[ |a_n| = |a_n-L+L| \le |a_n-L|+|L| < 1+|L| . ]

Thus every term after the (N)-th one satisfies

[ |a_n|\le 1+|L|\qquad (n\ge N). ]

Step 2 – bound the finitely many “initial’’ terms

The set ({\, a_1 , a_2 ,\dots, a_{N-1} \,}) contains only finitely many numbers, so it has a maximum (the real numbers are totally ordered and every finite non‑empty set has a maximum).

Define

[ M_0:=\max{|a_1|,|a_2|,\dots,|a_{N-1}|}. ]

If (N=1) this set is empty; in that case we may set (M_0:=0).

Step 3 – combine the two bounds

Let

[ M:=\max\bigl{M_0,\;1+|L|\bigr}. ]

Both (M_0) and (1+|L|) are real numbers, so their maximum exists.
Now we check the two possible cases for an arbitrary index (n).

If (n\ge N):
(|a_n|\le 1+|L|\le M) by the definition of (M).

If (n<N):
(|a_n|\le M_0\le M) again by the definition of (M).

Hence for every (n\in\mathbb N) we have (|a_n|\le M).
Therefore the sequence ((a_n)) is bounded.

3. Why the “maximum of all terms’’ idea does not work

Idea Why it is not a valid proof        
“Take (M=\max{ a_n :n\in\mathbb N}). If a maximum does not exist, take a number larger than that supposed maximum.” Circular reasoning: To speak of the maximum of all the absolute values we must already know that such a maximum exists. The existence of that maximum is exactly what we are trying to prove. Without a prior bound we cannot guarantee the set ({ a_n :n\in\mathbb N}) has a supremum that is finite, nor that a maximum exists.
“If there is no maximum, pick any number larger than the (non‑existent) maximum.” The phrase “larger than the non‑existent maximum’’ is meaningless: a number larger than something that does not exist cannot be defined. Moreover, we cannot simply choose a number larger than every term unless we have already shown that the collection of terms is bounded above.        
“Because the sequence converges, its terms must stay close to the limit, so the set of absolute values must have a maximum.” Convergence guarantees that the terms eventually lie inside any prescribed neighbourhood of the limit, but only after some index (N). The finitely many terms before (N) might be arbitrarily large; we need a separate argument (the finiteness argument above) to bound those. The existence of a global maximum is not automatic.        

In short, the “maximum of all terms’’ approach assumes the very property (boundedness) we are trying to establish, making the argument circular and invalid.

4. Final answer

A convergent sequence ((a_n)) of real numbers is always bounded.
A correct proof proceeds by:

  1. Using the definition of convergence with (\varepsilon=1) to bound all terms from some index (N) onward by (1+ L );
  2. Bounding the finitely many preceding terms by taking their maximum;
  3. Taking the larger of the two numbers as a global bound (M).
Thus there exists a real number (M) such that ( a_n \le M) for every (n).

5. Common Mistakes

Mistake Explanation    
Assuming a maximum exists before proving boundedness The set ({ a_n :n\in\mathbb N}) may have no maximum (e.g., (a_n=n)). Boundedness must be shown first.
Using “take a number larger than the maximum’’ when the maximum does not exist “Larger than the non‑existent maximum’’ is undefined; you need a concrete finite upper bound, which comes from the convergence argument.    
Ignoring the finitely many early terms Convergence only controls the tail of the sequence. Early terms can be arbitrarily large unless handled separately.    
Choosing (\varepsilon) that depends on the sequence The proof works for any fixed (\varepsilon>0); (\varepsilon=1) is a convenient choice. Using a value that itself depends on the unknown bound leads to circularity.    
Working with “supremum’’ without justification The supremum exists for any non‑empty set of real numbers, but it may be (+\infty). One must first prove the supremum is finite; otherwise it does not give a bound.    

Avoid these pitfalls by following the three‑step argument shown above.

Original question: Boundedness of convergent sequence. on Mathematics Stack Exchange, licensed CC BY-SA.

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