Polynomial Growth in the Coefficients

Why Polynomial Factors Matter Less Than They Seem

A coefficient such as \(n^4\) grows without bound, while \(1/n^4\) decays towards zero. It may seem that either factor should substantially change the convergence of a power series. For the radius of convergence, however, the decisive comparison is with the coefficient’s \(n\)th root. The \(n\)th root of either polynomial growth or polynomial decay tends to \(1\). Thus, these factors do not change the radius supplied by the rest of the coefficient sequence.

This does not mean that polynomial factors are irrelevant to every question about a series. They can affect convergence at a boundary point. Here we determine the radius, which describes the series’ behavior strictly inside and outside a particular distance from its center. The boundary itself requires separate tests.

The Root Formula for a Power Series

Write a power series centered at \(c\) as \(\sum_{n=0}^{\infty} a_n(x-c)^n\). Its radius of convergence \(R\) is given by the Cauchy–Hadamard formula. In this formula, the limit superior accounts for coefficient sequences whose \(n\)th roots do not converge.

Formula: For a power series with coefficients \(a_n\), the radius of convergence is determined by

\frac{1}{R}=\limsup_{n\to\infty}|a_n|^{1/n}.

Here \(1/0=\infty\) and \(1/\infty=0\). If \(|a_n|^{1/n}\) has a finite limit \(L>0\), then \(R=1/L\).

Consequently, if a coefficient is a product of a polynomial-size factor and another factor, we can examine their \(n\)th roots separately. For \(n\geq 1\) and a fixed real number \(\alpha\), the polynomial-size factor is \(n^\alpha\), including when \(\alpha\) is negative.

Why the \(n\)th Root Removes a Power of \(n\)

The key limit is \(n^{\alpha/n}\to 1\). For positive \(n\), its logarithm is \(\alpha\ln(n)/n\). To see that this tends to zero, use \(\ln(n)=\int_1^n 1/t\,dt\). Since \(1/t\leq 1/\sqrt{t}\) for \(t\geq1\), we have \(0\leq\ln(n)\leq2(\sqrt n-1)\). Dividing by \(n\) shows that \(0\leq\ln(n)/n\leq2/\sqrt n\), which tends to zero. Therefore \(\alpha\ln(n)/n\to0\), and exponentiating gives the stated limit.

\bigl(n^\alpha\bigr)^{1/n} =\exp\!\left(\frac{\alpha\ln n}{n}\right) \longrightarrow \exp(0)=1.

This argument applies to every fixed real \(\alpha\). For \(\alpha>0\), the coefficients \(n^\alpha\) grow; for \(\alpha<0\), they decay. In either case, taking the \(n\)th root makes their contribution tend to \(1\).

The Radius Is Unchanged by a Power of \(n\)

Suppose a coefficient sequence has the form \(a_n=n^\alpha b_n\), where \(\alpha\) is a fixed real number and \(|b_n|^{1/n}\to L\) for some finite \(L>0\). Then

|a_n|^{1/n} =\bigl(n^\alpha\bigr)^{1/n}|b_n|^{1/n} \longrightarrow 1\cdot L=L.

The Cauchy–Hadamard formula therefore gives \(R=1/L\). The polynomial factor changes the coefficients but not the radius. This conclusion requires that the remaining coefficient factor have the stated positive finite root limit. We cannot use this particular limit argument if that factor’s \(n\)th roots do not approach such an \(L\); the limit-superior formula remains the general radius formula.

Theorem: If \(a_n=n^\alpha b_n\), where \(\alpha\) is fixed and \(|b_n|^{1/n}\to L\) with \(0<L<\infty\), then the power series with coefficients \(a_n\) has radius \(R=1/L\). Multiplying the coefficients by a fixed power of \(n\) does not change the radius.

A geometric factor makes the result especially easy to use. If \(a_n=n^\alpha q^n\) with \(q\neq0\), then

|a_n|^{1/n}=n^{\alpha/n}|q| \longrightarrow |q|.

The radius is \(1/|q|\). A polynomial factor multiplying \(q^n\) does not alter that value.

Worked Examples

Worked Example: Coefficients Growing Like \(n^3\)

Find the radius of convergence of \(\sum_{n=1}^{\infty}n^3x^n\). Here the coefficient of \(x^n\) is \(a_n=n^3\). Its \(n\)th root is

|a_n|^{1/n}=(n^3)^{1/n}=n^{3/n}\longrightarrow1.

Thus \(1/R=1\), so \(R=1\). The check is that the root limit and the resulting radius are reciprocals: \(1/R=1/1=1\), as required. The radius tells us convergence for \(|x|<1\) and divergence for \(|x|>1\). It does not decide what happens at \(x=1\) or \(x=-1\).

Worked Example: Polynomial Decay with a Geometric Factor

Find the radius of \(\sum_{n=1}^{\infty}\frac{4^n}{n^2}(x+2)^n\). The series is centered at \(-2\), and its coefficient is \(a_n=4^n/n^2\). Take the \(n\)th root:

|a_n|^{1/n} =\left(\frac{4^n}{n^2}\right)^{1/n} =\frac{4}{n^{2/n}} \longrightarrow\frac{4}{1}=4.

Therefore \(1/R=4\), so \(R=1/4\). Checking the reciprocal gives \(1/R=1/(1/4)=4\), matching the root limit. The series converges for \(|x+2|<1/4\) and diverges for \(|x+2|>1/4\). This radius calculation does not determine convergence when \(|x+2|=1/4\).

Worked Example: A Polynomial in the Coefficients

Find the radius of \(\sum_{n=1}^{\infty}\frac{5n^2+3n+1}{7^n}x^n\). Its coefficient is \(a_n=(5n^2+3n+1)/7^n\). For \(n\geq1\), the numerator satisfies

5n^2\leq5n^2+3n+1\leq9n^2.

The lower bound holds because \(3n+1\geq0\); the upper bound follows from \(3n\leq3n^2\) and \(1\leq n^2\). Taking \(n\)th roots gives

5^{1/n}n^{2/n} \leq(5n^2+3n+1)^{1/n} \leq9^{1/n}n^{2/n}.

Both bounds tend to \(1\): the fixed positive constants have \(n\)th roots tending to \(1\), and \(n^{2/n}\to1\). The squeeze theorem therefore gives \((5n^2+3n+1)^{1/n}\to1\). It follows that

|a_n|^{1/n} =\frac{(5n^2+3n+1)^{1/n}}{7} \longrightarrow\frac17.

Hence \(R=7\). The check \(1/R=1/7\) agrees with the root limit. The series converges for \(|x|<7\) and diverges for \(|x|>7\); this result alone does not settle either boundary point.

Polynomial Factors More Generally

The same idea applies to a polynomial \(P(n)\), not only to a single power \(n^\alpha\). If \(P\) is a nonzero polynomial of degree \(d\), then its leading term has the form \(c n^d\) with \(c\neq0\). For sufficiently large \(n\), there are positive constants \(C_1,C_2\) such that \(C_1n^d\leq|P(n)|\leq C_2n^d\). Taking \(n\)th roots bounds \(|P(n)|^{1/n}\) between \(C_1^{1/n}n^{d/n}\) and \(C_2^{1/n}n^{d/n}\). Each bound tends to \(1\), so \(|P(n)|^{1/n}\to1\).

Thus a nonzero polynomial factor multiplying coefficients with a positive finite root limit leaves that limit, and therefore the radius, unchanged. The same is true for a reciprocal polynomial factor when it is defined for the indices in question: its \(n\)th root also tends to \(1\).

Common Pitfall: Confusing Coefficient Size with Root Growth

A common error is to reason that because \(n^3\) grows without bound, it must force a smaller radius of convergence. For the series \(\sum n^3x^n\), that mistaken reasoning might claim that the coefficients grow exponentially. But the root test uses the \(n\)th root of each coefficient, not the coefficient by itself.

\text{Incorrect comparison: } n^3\to\infty \quad\Longrightarrow\quad\text{“the radius must shrink.”}

The conclusion does not follow: unbounded growth alone does not determine the radius. The correct calculation is

(n^3)^{1/n}=n^{3/n}\longrightarrow1,

so the coefficient factor \(n^3\) contributes a root limit of \(1\). In this example, the radius is \(1\), not a smaller value inferred from coefficient growth alone.

What the Radius Does—and Does Not—Tell You

The radius summarizes convergence according to distance from the center: a power series converges absolutely when \(|x-c|<R\), and diverges when \(|x-c|>R\). Points satisfying \(|x-c|=R\) lie on the boundary. The root-limit calculation for polynomial factors finds \(R\), but it does not classify those boundary points. In particular, do not infer boundary convergence merely because a polynomial factor has \(n\)th root tending to \(1\). Substitute a boundary point into the series and examine the resulting series separately.

Key takeaway: For fixed \(\alpha\), \(\bigl(n^\alpha\bigr)^{1/n}\to1\). Therefore a factor \(n^\alpha\), or a nonzero polynomial factor, does not change a finite positive root limit supplied by the rest of the coefficients. Use that root limit to find the radius, and analyze boundary points separately.

Check Your Understanding

For each question, distinguish the coefficient’s ordinary size from its \(n\)th-root behavior.

What is \(\lim_{n\to\infty}(n^5)^{1/n}\), and why?
Find the radius of \(\sum_{n=1}^{\infty}n^2 3^n x^n\).
Find the radius of \(\sum_{n=1}^{\infty}n^{-3}(2x)^n\).
Why does multiplying coefficients by a fixed power of \(n\) leave a positive finite root limit unchanged?
If a root calculation gives a radius \(R\), what does it tell you about points satisfying \(|x-c|=R\)?