Oscillating Factors and Squeeze Bounds

When Oscillation Is Tamed by a Denominator

Some sequences have terms that continue to change direction or value as the index increases. That behavior does not prevent the sequence from converging if the size of each term is forced toward zero by a denominator that grows without bound.

For example, the sine values in \(\sin(n)\) oscillate as \(n\) ranges over the positive integers. Yet the sequence \(\sin(n)/n\) converges to zero. The key is not to determine a limit for \(\sin(n)\), but to use the fact that sine always lies between \(-1\) and \(1\). Dividing those bounds by \(n\) gives bounds that converge to zero.

Central idea: If an oscillating factor stays bounded while a positive denominator grows without bound, the quotient can be squeezed between bounds that tend to zero.

Bounded Factors and Squeeze Bounds

A sequence \((u_n)\) is bounded if there is a fixed number \(M\geq 0\) such that \(|u_n|\leq M\) for every \(n\in\mathbb{N}\). The value of \(u_n\) may change with \(n\); boundedness only says that its size never exceeds the same fixed ceiling.

The sine and cosine functions satisfy \(-1\leq\sin(x)\leq 1\) and \(-1\leq\cos(x)\leq 1\) for every real \(x\). Thus, for instance, \(u_n=\sin(n)\) is bounded by \(M=1\). The alternating factor \(u_n=(-1)^n\) is also bounded by \(1\), because its absolute value is \(1\) for every positive integer \(n\).

Theorem: If \(|u_n|\leq M\) for every \(n\in\mathbb{N}\), where \(M\) is a fixed nonnegative real number, then \(\lim_{n\to\infty}u_n/n=0\).

This result does not require \(u_n\) itself to converge. It requires only a uniform bound on its size. The proof is a direct application of the Squeeze Theorem for sequences, together with an explicit epsilon check.

Why the Bound Forces the Limit

From \(|u_n|\leq M\), we have \(-M\leq u_n\leq M\). Since \(n\) is positive, division by \(n\) preserves the inequality:

-\frac{M}{n}\leq\frac{u_n}{n}\leq\frac{M}{n}.

Both outside sequences tend to zero: \(M/n\to 0\) and \(-M/n\to 0\). The Squeeze Theorem therefore gives \(u_n/n\to 0\). The following epsilon argument spells out why the bounds suffice. Given any \(\varepsilon>0\), choose \(N\in\mathbb{N}\) such that \(N>M/\varepsilon\). For every \(n\geq N\),

\left|\frac{u_n}{n}-0\right| =\frac{|u_n|}{n} \leq\frac{M}{n} \leq\frac{M}{N} <\varepsilon.

This is precisely the definition of convergence to zero: for every positive tolerance \(\varepsilon\), all terms from some index onward lie within that tolerance of zero. If \(M=0\), then every \(u_n=0\), so the conclusion holds immediately as well.

Worked Example: The Sequence \(\sin(n)/n\)

Consider \(a_n=\sin(n)/n\), where \(n\in\mathbb{N}\). Since sine is bounded between \(-1\) and \(1\), we have \(-1\leq\sin(n)\leq 1\). Because \(n>0\), dividing through by \(n\) gives

-\frac{1}{n}\leq\frac{\sin(n)}{n}\leq\frac{1}{n}.

The two bounding sequences both converge to zero, so the Squeeze Theorem gives \(\lim_{n\to\infty}a_n=0\).

Check: For any \(\varepsilon>0\), choose \(N=\lfloor 1/\varepsilon\rfloor+1\). Then \(N>1/\varepsilon\), and for every \(n\geq N\),

|a_n-0| =\frac{|\sin(n)|}{n} \leq\frac{1}{n} \leq\frac{1}{N} <\varepsilon.

This verifies the claimed limit directly. The sine factor need not have a limit for this argument to work.

Absolute Values Handle Sign Changes

When a numerator changes sign, the inequality with symmetric bounds is often the simplest route. If \(|a_n|\leq c_n\), where \(c_n\geq 0\), then the definition of absolute value gives \(-c_n\leq a_n\leq c_n\). If both bounds tend to zero, the Squeeze Theorem applies. This method handles positive and negative terms together instead of splitting the sequence into cases.

Worked Example: An Alternating Factor Divided by \(n\)

Let \(b_n=(-1)^n/n\). For each positive integer \(n\), \(|(-1)^n|=1\). Therefore

|b_n| =\left|\frac{(-1)^n}{n}\right| =\frac{|(-1)^n|}{n} =\frac{1}{n}.

Equivalently, \(-1/n\leq b_n\leq 1/n\). Both bounds tend to zero, so \(\lim_{n\to\infty}b_n=0\).

Check: Given \(\varepsilon>0\), choose \(N=\lfloor 1/\varepsilon\rfloor+1\). For every \(n\geq N\),

|b_n-0| =\frac{1}{n} \leq\frac{1}{N} <\varepsilon.

For example, \(b_4=1/4\) and \(b_5=-1/5\). The signs differ, but both terms satisfy \(|b_n|\leq 1/n\), the bound that establishes convergence.

Using a Bound That Fits the Whole Numerator

Sometimes the oscillating part is a combination of terms rather than one familiar bounded factor. The triangle inequality, \(|x+y|\leq|x|+|y|\), lets us build a fixed bound from known bounds. Once the numerator is bounded, division by a growing positive denominator can make the entire quotient small.

Worked Example: A Bounded Sine-Cosine Combination

Consider \(c_n=(4\sin(n)-3\cos(n))/(5n)\). The sine and cosine bounds give \(|\sin(n)|\leq 1\) and \(|\cos(n)|\leq 1\). Applying the triangle inequality shows

|4\sin(n)-3\cos(n)| \leq 4|\sin(n)|+3|\cos(n)| \leq 4(1)+3(1) =7.

Since \(5n>0\), it follows that

|c_n| =\frac{|4\sin(n)-3\cos(n)|}{5n} \leq\frac{7}{5n}.

Thus \(-7/(5n)\leq c_n\leq 7/(5n)\), and both bounds tend to zero. The Squeeze Theorem gives \(\lim_{n\to\infty}c_n=0\).

Check: Given \(\varepsilon>0\), choose \(N=\lfloor 7/(5\varepsilon)\rfloor+1\). Then \(N>7/(5\varepsilon)\). For \(n\geq N\),

|c_n-0| \leq\frac{7}{5n} \leq\frac{7}{5N} <\varepsilon.

The fixed numerator bound \(7\) is enough; its exact smallest possible value is not needed.

A Positive Oscillating Numerator

If a numerator has a known lower bound as well as an upper bound, the squeeze may use those bounds directly. For instance, \(1\leq 2+\sin(n^2)\leq 3\), since \(-1\leq\sin(n^2)\leq 1\). This produces a nonnegative sequence squeezed between zero and a quantity tending to zero.

Worked Example: A Shifted Sine Factor

Let \(d_n=(2+\sin(n^2))/(3n+1)\). The sine bound gives

-1\leq\sin(n^2)\leq 1 \quad\Longrightarrow\quad 1\leq 2+\sin(n^2)\leq 3.

Also \(3n+1>0\) for every \(n\in\mathbb{N}\). Dividing the numerator bounds by this positive denominator gives

0\leq d_n\leq\frac{3}{3n+1}.

For \(n>0\), \(3n+1>3n\), so \(3/(3n+1)<3/(3n)=1/n\). Hence

0\leq d_n<\frac{1}{n}.

The two outer bounds converge to zero, so \(\lim_{n\to\infty}d_n=0\).

Check: Given \(\varepsilon>0\), choose \(N=\lfloor1/\varepsilon\rfloor+1\). For every \(n\geq N\),

|d_n-0|=d_n<\frac{1}{n}\leq\frac{1}{N}<\varepsilon.

The lower bound \(d_n\geq0\) justifies replacing \(|d_n|\) by \(d_n\) in this check.

A Common Mistake: Assuming the Oscillating Factor Converges

A tempting but incorrect argument is to try to find a limit for the oscillating factor first. For example, one might write \((-1)^n\to-1\), then conclude that \((-1)^n/n\) behaves like \(-1/n\). The first statement is false: \((-1)^n\) is not always \(-1\), so it cannot be used as a limit calculation.

The correction is to avoid assigning the factor a limit. The valid calculation is

\left|\frac{(-1)^n}{n}\right| =\frac{|(-1)^n|}{n} =\frac{1}{n} \longrightarrow 0.

More generally, a bounded factor is not automatically convergent. The squeeze argument needs a bound on its magnitude and bounds for the whole quotient that converge to the same value. It does not justify a limit for the factor on its own.

When This Technique Is Useful

This method is useful whenever a sequence combines persistent variation with a shrinking scale. Rather than trying to track every oscillation, find a fixed bound for the part that varies and then compare the full term with a simple sequence whose limit is known. The result applies to sine, cosine, alternating signs, and any other bounded sequence in the numerator.

Before applying the squeeze, check the conditions carefully. The bound on the oscillating factor must be independent of \(n\), and the denominator must be positive so dividing preserves the order of inequalities. Finally, the outside bounds must converge to the same number. If any of these points is missing, the squeeze argument has not yet been established.

Takeaway: To show that a bounded oscillation divided by \(n\) tends to zero, bound its absolute value by \(M/n\). This bound tends to zero, regardless of whether the oscillating factor itself has a limit.

Check Your Understanding

Use bounds on the oscillating factor and the Squeeze Theorem where appropriate.

Find a fixed bound for the absolute value of \(\cos(n)/n\), then state its limit.
Show that \((3\sin(n)-2)/n\) converges to zero by bounding the numerator.
For \((1+\cos(n))/(2n)\), find lower and upper bounds that have the same limit.
Why is it unnecessary to prove that \((-1)^n\) converges before analyzing \((-1)^n/n\)?
State the two key conditions needed to divide an inequality by a denominator without changing its direction.