Exponential Decay Versus Polynomial Growth

Introduction

A sequence can combine two competing effects. A power such as $n^k$ grows as $n$ increases, while an exponential factor $r^n$, with $0<r<1$, decreases. Looking at either factor alone does not determine what happens to their product.

The key is to compare consecutive terms. For a fixed nonnegative integer $k$, the ratio between consecutive terms approaches $r$. Since $r<1$, the terms eventually shrink by at least a fixed factor less than one. Repeated shrinking gives a geometric bound, which proves that the sequence tends to zero.

Central result: If $k$ is a fixed nonnegative integer and $0<r<1$, then $n^k r^n$ tends to zero as $n$ tends to infinity.

Set Up the Sequence

Throughout this tutorial, $n$ is a positive integer, $k$ is a fixed nonnegative integer, and $r$ is a fixed real number satisfying $0<r<1$. Define

$$ a_n=n^k r^n. $$

The condition on $r$ means that multiplying by $r$ repeatedly produces exponential decay. The exponent $k$ is fixed: it does not change with $n$. If $k=0$, then $a_n=r^n$. If $k>0$, the power $n^k$ grows, so the result is not immediate from the separate behavior of the factors.

In particular, the product limit law does not settle the case $k>0$: one factor grows without bound and the other tends to zero. We need to compare their rates of change.

The Consecutive-Term Ratio

Since every $a_n$ is positive, we can divide consecutive terms. Using the definition of $a_n$,

\frac{a_{n+1}}{a_n} = \frac{(n+1)^k r^{n+1}}{n^k r^n} = \left(\frac{n+1}{n}\right)^k r = r\left(1+\frac{1}{n}\right)^k.

For fixed $k$, the factor $\left(1+\frac{1}{n}\right)^k$ tends to $1$. For $k>0$, this follows by expanding the finite power using the binomial theorem: all terms other than the constant term contain a factor $1/n^j$ for some positive integer $j$, and therefore tend to zero. For $k=0$, the factor is exactly $1$. Thus

\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=r.

This ratio limit is useful because $r<1$. Choose a number $q$ strictly between $r$ and $1$, for example $q=(1+r)/2$. Since the consecutive-term ratio tends to $r<q$, there is an index $N$ such that for every $n\geq N$,

\frac{a_{n+1}}{a_n}\leq q<1.

Multiplying these inequalities from $n=N$ through $n=m-1$ gives, for every $m\geq N$,

a_m = a_N\prod_{n=N}^{m-1}\frac{a_{n+1}}{a_n} \leq a_N q^{m-N}.

The product on the left telescopes: each intermediate term in a denominator cancels with the same term in a numerator, leaving $a_m/a_N$. Since $0<q<1$, the geometric factor $q^{m-N}$ tends to zero. For completeness, write $1/q=1+c$, where $c>0$. The binomial theorem gives $(1+c)^j\geq 1+jc$ for every nonnegative integer $j$, so $q^j\leq 1/(1+jc)$, which tends to zero. The bound therefore forces $a_m$ to tend to zero as well.

Theorem: For every fixed nonnegative integer $k$ and every real number $r$ with $0<r<1$, $$ \lim_{n\to\infty} n^k r^n=0. $$ The proof works by showing that the ratio of consecutive terms is eventually bounded above by a fixed number $q<1$, and then comparing the sequence with a geometric sequence.

Worked Examples

Worked Example: A Linear Power with Base One-Half

Find the limit of $a_n=n(1/2)^n$. First compute the ratio of consecutive terms:

\frac{a_{n+1}}{a_n} = \frac{(n+1)(1/2)^{n+1}}{n(1/2)^n} = \frac{n+1}{2n} = \frac{1}{2}+\frac{1}{2n}.

For $n\geq2$, $1/(2n)\leq1/4$, so the ratio is at most $3/4$. Also,

a_2=2\left(\frac{1}{2}\right)^2 =2\cdot\frac{1}{4} =\frac{1}{2}.

Repeatedly applying the ratio bound gives, for $n\geq2$,

0<a_n\leq a_2\left(\frac{3}{4}\right)^{n-2} = \frac{1}{2}\left(\frac{3}{4}\right)^{n-2}.

The right-hand side tends to zero, so the squeeze theorem gives $\lim_{n\to\infty}n(1/2)^n=0$. The calculation is checked by the valid bound $1/(2n)\leq1/4$ for all $n\geq2$, and by the exact starting value $a_2=1/2$.

Worked Example: A Quadratic Power with Base One-Third

Find the limit of $b_n=n^2(1/3)^n$. Its consecutive-term ratio is

\frac{b_{n+1}}{b_n} = \frac{(n+1)^2(1/3)^{n+1}}{n^2(1/3)^n} = \frac{1}{3}\left(\frac{n+1}{n}\right)^2 = \frac{1}{3}\left(1+\frac{1}{n}\right)^2.

For $n\geq4$, $1+1/n\leq5/4$. Therefore,

\frac{b_{n+1}}{b_n} \leq \frac{1}{3}\left(\frac{5}{4}\right)^2 = \frac{25}{48} < \frac{3}{5},

where the final comparison holds because $25\cdot5=125<144=3\cdot48$. The starting value is

b_4 = 4^2\left(\frac{1}{3}\right)^4 = 16\cdot\frac{1}{81} = \frac{16}{81}.

Consequently, for $n\geq4$,

0<b_n\leq\frac{16}{81}\left(\frac{3}{5}\right)^{n-4}.

Because the geometric factor tends to zero, the sequence tends to zero. The bound is valid from $n=4$ onward because the ratio estimate holds for every such $n$, and the stated starting value is obtained directly from the definition.

Worked Example: A Cubic Power with Base Two-Thirds

Find the limit of $c_n=n^3(2/3)^n$. The ratio is

\frac{c_{n+1}}{c_n} = \frac{(n+1)^3(2/3)^{n+1}}{n^3(2/3)^n} = \frac{2}{3}\left(1+\frac{1}{n}\right)^3.

For $n\geq10$, $1+1/n\leq11/10$. Hence

\frac{c_{n+1}}{c_n} \leq \frac{2}{3}\left(\frac{11}{10}\right)^3 = \frac{2}{3}\cdot\frac{1331}{1000} = \frac{1331}{1500} < \frac{9}{10}.

The last inequality follows from $1331\cdot10=13310<13500=9\cdot1500$. Calculate the starting value:

c_{10} = 10^3\left(\frac{2}{3}\right)^{10} = 1000\cdot\frac{1024}{59049} = \frac{1024000}{59049}.

For every $n\geq10$, the ratio bound gives

0<c_n\leq \frac{1024000}{59049}\left(\frac{9}{10}\right)^{n-10}.

The constant is large, but it is fixed; the geometric factor tends to zero. Thus $\lim_{n\to\infty}n^3(2/3)^n=0$. The estimate and starting term have both been computed exactly, so this bound verifies the limit.

Why the Ratio Argument Is Needed

It can be tempting to look at the two factors separately and conclude that their product tends to zero because $r^n$ tends to zero. That reasoning is incomplete when $k>0$: the other factor $n^k$ grows, and the product limit law does not apply to a factor that does not have a finite limit.

The ratio method handles the competition directly. Once the ratio of consecutive terms is at most $q<1$, every subsequent term is at most $q$ times the previous term. After $j$ steps, the term is bounded by its starting value times $q^j$. This is the precise reason that the decreasing exponential eventually outweighs the fixed power.

The conditions matter. When $r=1$, the sequence is $n^k$, which does not tend to zero for $k\geq0$. When $r>1$, the exponential factor does not decay. The theorem concerns a fixed $r$ strictly between zero and one; it does not assert the same conclusion for these other cases.

Common Pitfall: Treating the Ratio Limit as the Sequence Limit

For $a_n=n^2(1/2)^n$, the ratio calculation gives

\lim_{n\to\infty}\frac{a_{n+1}}{a_n} = \lim_{n\to\infty}\frac{1}{2}\left(1+\frac{1}{n}\right)^2 = \frac{1}{2}.

A mistaken conclusion would be: “Therefore, $a_n$ tends to $1/2$.” The ratio is not the sequence term; it only describes how consecutive terms compare. The correct conclusion follows by choosing a number such as $q=3/4$, which is larger than $1/2$ but still less than one. The ratio is eventually at most $3/4$, so the terms are bounded by a constant times a power of $3/4$, and that bound tends to zero.

The distinction is important: the ratio limit supplies a shrinking factor, while the geometric comparison establishes the limit of the sequence itself.

Summary

For $a_n=n^k r^n$, with fixed $k$ a nonnegative integer and $0<r<1$, the consecutive-term ratio is

\frac{a_{n+1}}{a_n} = r\left(1+\frac{1}{n}\right)^k \longrightarrow r<1.

It follows that the ratio is eventually at most some fixed $q<1$. From that point onward, the sequence is bounded by a constant times a geometric sequence with ratio $q$, which tends to zero. Therefore the original sequence tends to zero as well.

Check Your Understanding

Use consecutive-term ratios and geometric bounds where appropriate.

For $a_n=n^4(1/2)^n$, calculate $a_{n+1}/a_n$ and state its limit.
Why does a ratio limit strictly less than one allow a geometric bound?
Find an index from which the ratio for $n^2(1/4)^n$ is at most $1/2$.
What is the limit of $n^3(3/5)^n$, and which condition on the base allows the theorem to apply?
For $n^2(1/2)^n$, explain why its ratio limit is not the limit of the sequence.