Why Squeezing Helps
Some sequence limits are difficult to calculate directly. It can be easier to compare the sequence with two simpler ones: a lower bound and an upper bound. If both bounds approach the same number, the sequence trapped between them has no room to approach anything else.
The key is to make the comparison precise. The inequalities must hold for every sufficiently large index, and the two bounding sequences must converge to the same limit. The squeeze theorem turns those facts into a conclusion about the sequence between them.
Order and Convergence
A sequence \((a_n)\) converges to \(L\) if, for every \(\varepsilon>0\), there is an index \(N\in\mathbb{N}\) such that \(n\ge N\) implies \(|a_n-L|<\varepsilon\). In words, all sufficiently late terms lie within any chosen positive distance of \(L\).
For the squeeze theorem, we compare three real sequences \((b_n)\), \((a_n)\), and \((c_n)\). The sequence \((a_n)\) is squeezed between the other two when \(b_n\le a_n\le c_n\). These inequalities need only hold eventually: a finite number of early terms do not affect convergence.
The two bounding sequences must approach the same limit. It is not enough that each bound converges to some number, and the theorem does not say that a sequence between bounds with different limits must converge.
Why the Theorem Works
The proof uses the definition of convergence and the order of real numbers. Fix any tolerance \(\varepsilon>0\). Since \(b_n\to L\), the terms \(b_n\) eventually lie above \(L-\varepsilon\). Since \(c_n\to L\), the terms \(c_n\) eventually lie below \(L+\varepsilon\). Once both facts and the squeeze inequalities hold, \(a_n\) lies between those same two limits.
There is an index \(N_b\) such that \(n\ge N_b\) implies \(|b_n-L|<\varepsilon\). In particular, \(L-\varepsilon<b_n\).
There is an index \(N_c\) such that \(n\ge N_c\) implies \(|c_n-L|<\varepsilon\). In particular, \(c_n<L+\varepsilon\).
Let \(N=\max\{N_0,N_b,N_c\}\). For every \(n\ge N\), the squeeze inequalities and both convergence bounds hold.
For \(n\ge N\), \(L-\varepsilon<b_n\le a_n\le c_n<L+\varepsilon\). Thus \(|a_n-L|<\varepsilon\), as required.
Because this argument works for every \(\varepsilon>0\), it proves \(a_n\to L\) by the definition of convergence. Notice that the proof needs the inequalities only from \(N_0\) onward; the chosen index \(N\) accounts for that condition as well as for the two bounds.
Worked Applications
Worked Example: A Rational Sequence Converging to Zero
Consider \(a_n=\frac{n}{n^2+1}\), for \(n\in\mathbb{N}\). Its terms are nonnegative. Also \(n^2+1\ge n^2>0\), so dividing the positive numerator \(n\) by these positive denominators gives
The lower bound is the constant sequence \(0\), which converges to \(0\). The upper bound \(1/n\) also converges to \(0\): given \(\varepsilon>0\), choose \(N>1/\varepsilon\); then \(n\ge N\) implies \(0<1/n<\varepsilon\). The squeeze theorem therefore gives
Check: for every positive integer \(n\), the denominator \(n^2+1\) is positive, and the two displayed inequalities hold. The bounds have the same limit, so all the theorem's conditions are satisfied.
Worked Example: A Rational Sequence Converging to One
Let \(a_n=\frac{n^2}{n^2+n+1}\). The denominator exceeds the numerator by \(n+1>0\), so \(a_n\le1\). To find a lower bound, calculate the difference from \(1-\frac1n\):
Here the numerator in the final simplification follows from \((n-1)(n^2+n+1)=n^3-1\). Thus, for every \(n\in\mathbb{N}\),
The lower bound tends to \(1\), since \(1/n\to0\), and the upper bound is the constant sequence \(1\). The squeeze theorem yields
Check: the calculated difference is positive, confirming the lower inequality, and the denominator \(n^2+n+1\) is at least \(n^2\), confirming the upper inequality. Both bounds approach \(1\), as required.
Worked Example: A Difference of Square Roots
For \(n\in\mathbb{N}\), consider \(a_n=\sqrt{n^2+4n}-\sqrt{n^2+n}\). Both radicands are positive, and the first is larger, so \(a_n>0\). Rationalizing the difference gives
For \(x\ge0\), \(1\le\sqrt{1+x}\le1+x/2\). The lower inequality follows from \(1+x\ge1\). For the upper inequality, both sides are nonnegative and \[ (1+x/2)^2=1+x+x^2/4\ge1+x, \] so taking square roots gives the claim. Apply these bounds to the denominator above:
All quantities are positive. Taking reciprocals reverses the inequalities, and multiplying by \(3\) gives
The upper bound is the constant sequence \(3/2\). For the lower bound,
so \(6n/(4n+5)\to3/2\). The squeeze theorem now proves
Check: rationalization is valid because the sum of the two square roots is positive. The bounds were obtained from positive quantities, and both approach \(3/2\), matching the theorem's conditions.
Choosing Bounds and Avoiding a Common Error
A useful bound is one that is simpler to analyze than the original sequence. Nonnegative expressions can often be bounded above by a simpler positive expression; a sequence near a proposed limit can sometimes be bounded below and above by expressions that both approach that limit. The inequalities do not need to be strict, and they do not need to hold for every index if they hold for all sufficiently large indices.
A common mistake is to assume that merely having some lower and upper bounds determines a limit. For example, \(0\le a_n\le1\) does not imply \(a_n\to0\). The constant sequence \(a_n=1/2\) satisfies both inequalities for every \(n\), but it converges to \(1/2\), not \(0\). The bounds in this example approach different numbers, so the squeeze theorem cannot identify a single limit.
The theorem is a sufficient condition, not a required method for every limit. If direct algebra or a limit law already gives the answer, squeezing may not be necessary. Its particular strength is that it can establish a limit without calculating the middle sequence's terms in a simpler exact form.
Check Your Understanding
For each question, identify the inequalities and limits that justify the conclusion, or explain why the squeeze theorem does not apply.
- If \(b_n\le a_n\le c_n\) for all \(n\ge 7\), and both \(b_n\) and \(c_n\) converge to \(4\), what can be concluded about \(a_n\)?
- For \(a_n=\frac{n}{n^2+3}\), find a simple nonnegative upper bound that converges to zero.
- Why is it valid for the squeeze inequalities to begin only at some index \(N_0\), rather than holding for every \(n\in\mathbb{N}\)?
- Does \(0\le a_n\le2\) prove that \(a_n\) converges? Explain what additional information the squeeze theorem requires.
- In the square-root example, why does taking reciprocals reverse the inequalities for the denominator bounds?
