Logarithmic Growth Versus Power Growth

Comparing Two Slow-Growth Patterns

Both the natural logarithm $\ln n$ and a positive power $n^p$ increase as $n$ increases. Their rates of increase, however, are very different. A logarithm grows more slowly than every positive power: no matter how small $p$ is, $n^p$ eventually outgrows $\ln n$.

This comparison gives two useful conclusions. The logarithm divided by a positive power tends to zero, while the power divided by the logarithm grows without bound. The proof below uses intervals between consecutive powers of $2$, together with a geometric-sequence estimate.

Notation: Throughout, $\ln x$ denotes the natural logarithm for $x>0$, and $p$ is a fixed positive real number. We use that $\ln x$ is increasing, $\ln(ab)=\ln a+\ln b$ for $a,b>0$, and $n^p$ is positive for $n\geq 1$.

The Main Comparison

Theorem: For every fixed real number $p>0$, $$ \lim_{n\to\infty}\frac{\ln n}{n^p}=0 \qquad\text{and}\qquad \lim_{n\to\infty}\frac{n^p}{\ln n}=+\infty. $$ The second limit is understood for $n\geq 2$, where $\ln n>0$. A sequence tends to $+\infty$ when, for every $M>0$, all sufficiently late terms are greater than $M$.

The first limit says that the logarithm becomes negligible relative to any fixed positive power. The second is its reciprocal comparison: the power eventually exceeds the logarithm by any prescribed factor. We prove the first limit and then justify the reciprocal conclusion.

Why the Comparison Holds

For each integer $n\geq 1$, choose the integer $k\geq 0$ satisfying $2^k\leq n<2^{k+1}$. Such an integer exists because the powers of $2$ increase without bound. Since the logarithm is increasing, and since $n^p$ is increasing for $p>0$, these inequalities give bounds on the numerator and denominator separately:

\ln n\leq \ln(2^{k+1})=(k+1)\ln 2, \qquad n^p\geq (2^k)^p=2^{kp}.

Thus the ratio is nonnegative and satisfies

0\leq\frac{\ln n}{n^p} \leq (\ln 2)\frac{k+1}{2^{kp}}.

It remains to show that the right-hand side tends to zero as $k$ increases. Set $a=2^p$, so $a>1$, and define $b_k=(k+1)/a^k$. The ratio of consecutive terms is

\frac{b_{k+1}}{b_k} = \frac{k+2}{a(k+1)} = \frac{1+\frac{1}{k+1}}{a}.

Choose a number $r$ with $1/a<r<1$. Since $ar>1$, there is an integer $K$ large enough that $1+1/(K+1)\leq ar$. For every $k\geq K$, it follows that $b_{k+1}/b_k\leq r$. Repeatedly applying this inequality gives $b_k\leq b_Kr^{k-K}$ for $k\geq K$. Because $0<r<1$, the geometric sequence $r^{k-K}$ tends to zero, so $b_k\to 0$.

As $n$ increases without bound, its associated index $k$ also increases without bound: for any fixed $K$, $n\geq 2^K$ ensures that $k\geq K$. Therefore the upper bound $(\ln 2)b_k$ tends to zero as $n\to\infty$. The squeeze theorem now gives the first limit.

Place $n$ in a dyadic interval.
Choose $k$ so that $2^k\leq n<2^{k+1}$.

Bound the logarithm and power.
Use $\ln n\leq(k+1)\ln 2$ and $n^p\geq 2^{kp}$.

Control the remaining factor.
The sequence $(k+1)/2^{kp}$ tends to zero by a geometric-ratio estimate.

Apply the squeeze theorem.
The original nonnegative ratio is bounded above by a sequence tending to zero.

To obtain the second limit, let $M>0$. The first limit implies that, for all sufficiently large $n$, $0<\ln n/n^p<1/M$. Taking reciprocals of positive quantities reverses the inequality, giving $n^p/\ln n>M$. This is exactly the definition of divergence to $+\infty$.

Worked Examples

Worked Example: A Logarithm Divided by a Cube Root

Evaluate $\lim_{n\to\infty}\ln n/n^{1/3}$. Here $p=1/3>0$, so the theorem applies. To see the bound explicitly, choose $k$ with $2^k\leq n<2^{k+1}$. Then

0\leq\frac{\ln n}{n^{1/3}} \leq\frac{(k+1)\ln 2}{2^{k/3}} =(\ln 2)\frac{k+1}{(2^{1/3})^k}.

The last factor tends to zero by the geometric-ratio argument with $a=2^{1/3}>1$. The lower bound is zero, so the squeeze theorem gives the limit. Check: the expression is nonnegative for $n\geq1$, and the displayed upper bound tends to zero; hence its limit is indeed zero.

Worked Example: A Power Divided by a Logarithm

Evaluate the behavior of $n^{2/3}/\ln n$ as $n\to\infty$. For $n\geq2$, both $n^{2/3}$ and $\ln n$ are positive. The first part of the theorem, with $p=2/3$, gives

\lim_{n\to\infty}\frac{\ln n}{n^{2/3}}=0.

More explicitly, for any chosen $M>0$, there is an index $N$ such that $n\geq N$ implies $0<\ln n/n^{2/3}<1/M$. Taking reciprocals yields $n^{2/3}/\ln n>M$. Since this works for every $M>0$,

\lim_{n\to\infty}\frac{n^{2/3}}{\ln n}=+\infty.

Check: the reciprocal inequality holds because both quantities being inverted are positive for $n\geq2$; it confirms divergence to $+\infty$, not convergence to a finite number.

Worked Example: A Logarithm with a Polynomial Input

Evaluate $\lim_{n\to\infty}\ln(n^2+1)/\sqrt n$. For $n\geq1$, $n^2+1\leq2n^2$. Monotonicity and the product rule for logarithms give

0\leq\ln(n^2+1) \leq\ln(2n^2) =\ln 2+2\ln n.

Divide all terms by $\sqrt n>0$:

0\leq\frac{\ln(n^2+1)}{\sqrt n} \leq\frac{\ln 2}{\sqrt n} +2\frac{\ln n}{n^{1/2}}.

The first term on the right tends to zero because it is a constant divided by $\sqrt n$. The second tends to zero by the theorem with $p=1/2$. Thus the entire upper bound tends to zero, and the squeeze theorem gives

\lim_{n\to\infty}\frac{\ln(n^2+1)}{\sqrt n}=0.

Check: the original ratio is nonnegative, and the stated upper bound tends to zero, verifying the limit.

Logarithm Bases and Constant Factors

The theorem is stated using the natural logarithm, but changing the base only multiplies a logarithm by a fixed positive constant. For a fixed base $b>1$, the change-of-base identity is $\log_b n=\ln n/\ln b$, where $\ln b>0$. Therefore

\frac{\log_b n}{n^p} =\frac{1}{\ln b}\frac{\ln n}{n^p} \longrightarrow 0, \qquad \frac{n^p}{\log_b n} =(\ln b)\frac{n^p}{\ln n} \longrightarrow+\infty.

The same comparison remains useful when a logarithm has a more complicated positive input. The key is to bound that input in a way that reduces it to a constant plus a fixed multiple of $\ln n$, as in the third example. The constants do not change the zero limit after division by $n^p$.

A Common Mistake

A tempting but incorrect manipulation is to claim that $\ln n/n^p=\ln(n/n^p)$. The logarithm does not distribute over division in this way: its rule is $\ln(a/b)=\ln a-\ln b$, not $\ln a/b=\ln(a/b)$. For example, when $n=4$ and $p=1$, the left side is $\ln 4/4$, while the proposed right side is $\ln 1=0$; these are not equal.

The correct comparison does not combine the quotient with the logarithm. Instead, it bounds $\ln n$ on dyadic intervals and compares that bound with $n^p$, as in the proof. Another error is to conclude that $n^p/\ln n$ tends to zero merely because the denominator grows. Both numerator and denominator grow; their relative rates matter. The reciprocal argument above shows that this particular quotient tends to $+\infty$.

Why This Growth Comparison Matters

In sequence-limit problems, terms may involve logarithms and powers at the same time. Knowing which grows faster lets you identify the dominant factor without treating the two divergent quantities as if they had the same size. In particular, any fixed positive power eventually dominates a logarithmic factor, however small the power is.

The condition $p>0$ is essential to this result: it makes $2^p>1$, which is what allows the geometric bound in the proof to decay. The theorem makes no claim for $p=0$, when $n^p=1$, or for negative $p$. Check the exponent before applying the comparison.

Check Your Understanding

Use the theorem and its proof to answer these questions.

What is $\lim_{n\to\infty}\ln n/n^{1/10}$, and which condition on the exponent lets you apply the theorem?
Explain why $n^{3/5}/\ln n$ tends to $+\infty$, rather than to zero.
For $n\geq1$, use $n^3+1\leq2n^3$ to find the limit of $\ln(n^3+1)/n^{1/4}$.
In the dyadic-interval proof, why does $(k+1)/2^{kp}$ tend to zero when $p>0$?
Why is the equality $\ln n/n^p=\ln(n/n^p)$ invalid?