Geometric Coefficients Set a Natural Scale
A power series is strongly shaped by how quickly its coefficients grow. When coefficients follow a geometric pattern, the radius of convergence can be found by comparing that pattern directly with the powers of the variable. Polynomial factors such as \(n^2\) or \(n^3+2n\) may change the size of individual coefficients, but they do not change the radius set by the geometric part.
Throughout this tutorial, take \(q>0\), and consider a power series centered at \(c\), written as \(\sum_{n=0}^{\infty} a_n(x-c)^n\). The radius describes the distance from \(c\) within which the series converges absolutely. The behavior exactly on the boundary is a separate question.
Pure Geometric Coefficients
For coefficients \(a_n=q^n\), the \(n\)th term of the power series is \(q^n(x-c)^n\). This is a geometric sequence in \(n\), with common ratio \(q(x-c)\). The geometric series converges when the absolute value of its common ratio is less than \(1\), so it converges when \(q|x-c|<1\). Equivalently, \(|x-c|<1/q\).
This agrees with the ratio-test calculation. If \(x\ne c\), the ratio of the absolute values of consecutive terms is \(q|x-c|\). When this is less than \(1\), the ratio test gives absolute convergence; when it is greater than \(1\), the terms do not tend to zero and the series diverges. The boundary occurs at distance \(1/q\) from \(c\).
For coefficients \(a_n=q^{-n}=1/q^n\), the \(n\)th term is \(\bigl((x-c)/q\bigr)^n\). Its geometric ratio has absolute value \(|x-c|/q\), so the series converges when \(|x-c|<q\). Thus its radius is \(q\).
These formulas express the same principle in two forms: the radius is the reciprocal of the geometric growth rate in the coefficients. Larger \(q\) makes \(q^n\) grow faster and reduces the radius. In contrast, larger \(q\) makes \(q^{-n}\) decay faster and increases the radius.
Worked Example: Coefficients \((3/5)^n\)
Worked Example: Coefficients \((3/5)^n\)
Find the radius of the series \(\sum_{n=0}^{\infty}(3/5)^n x^n\), centered at \(0\).
Here \(q=3/5\), so the terms can be written as a geometric series with ratio \((3/5)x\). The convergence condition is that the absolute value of this ratio be less than \(1\):
Therefore the radius is \(R=5/3\). As a check, at \(x=1\) the absolute ratio is \(3/5<1\), so the series converges absolutely. At \(x=2\), the absolute ratio is \(6/5>1\); the terms do not tend to zero, so the series diverges. These checks lie respectively inside and outside the calculated radius.
Polynomial Factors Do Not Change the Exponential Scale
Now consider coefficients \(P(n)q^n\), where \(P\) is a nonzero polynomial and \(q>0\). A polynomial can make the coefficients larger or smaller by a factor that depends on \(n\), but its growth is slower than any fixed exponential growth. The root test makes this precise.
Suppose \(P\) has degree \(d\) and leading coefficient \(A\ne0\). For all sufficiently large \(n\), its leading term determines its size: there are positive constants \(C_1,C_2\) such that \(C_1n^d\le |P(n)|\le C_2n^d\). Taking \(n\)th roots shows that \(|P(n)|^{1/n}\) tends to \(1\), because \(C_1^{1/n}\) and \(C_2^{1/n}\) tend to \(1\), and \(n^{d/n}\) tends to \(1\). Therefore the polynomial contributes no lasting factor to the coefficient growth measured by \(n\)th roots.
For \(a_n=P(n)q^n\), take \(n\)th roots to obtain \(|a_n|^{1/n}=|P(n)|^{1/n}q\).
Since \(|P(n)|^{1/n}\to1\), the coefficient-root limit is \(q\).
The radius is \(R=1/q\), just as for the pure geometric coefficients \(q^n\).
The same reasoning applies to \(P(n)q^{-n}\): the coefficient-root limit is \(1/q\), so the radius is \(q\). In both cases the polynomial changes the coefficients’ size, but not their exponential scale.
This statement also covers familiar factors such as \(n^k\), where \(k\) is a nonnegative integer. For \(k=0\), the factor is \(1\). For positive \(k\), the coefficient at \(n=0\) may be zero; changing or omitting a finite number of initial coefficients cannot affect the radius.
Worked Example: A Polynomial Times \(2^n\)
Worked Example: A Polynomial Times \(2^n\)
Find the radius of \(\sum_{n=0}^{\infty}(n+1)^2 2^n x^n\), centered at \(0\). The coefficients combine a polynomial factor \((n+1)^2\) with \(q^n\), where \(q=2\). The theorem predicts \(R=1/2\). Verify this with the ratio test.
Let \(u_n=(n+1)^2 2^n x^n\). For \(x\ne0\), the ratio of consecutive absolute term values is
Since \((n+2)/(n+1)\to1\), this ratio tends to \(2|x|\). The ratio test gives convergence when \(2|x|<1\), or \(|x|<1/2\), and divergence when \(2|x|>1\), or \(|x|>1/2\). Thus \(R=1/2\).
Check one point on each side. At \(x=1/4\), the limiting ratio is \(2(1/4)=1/2<1\), so the series converges absolutely. At \(x=3/4\), the limiting ratio is \(2(3/4)=3/2>1\), so it diverges. Both results agree with the radius.
Worked Example: A Polynomial Times \(4^{-n}\)
Worked Example: A Polynomial Times \(4^{-n}\)
Find the radius of \(\sum_{n=0}^{\infty}(n+1)^3 4^{-n}x^n\), centered at \(0\). Since \(4^{-n}=q^{-n}\) with \(q=4\), the predicted radius is \(R=4\). We can check it directly.
Let \(v_n=(n+1)^3 4^{-n}x^n\). For \(x\ne0\),
The factor \((n+2)/(n+1)\) tends to \(1\), so the ratio tends to \(|x|/4\). The ratio test gives convergence if \(|x|/4<1\), which is \(|x|<4\), and divergence if \(|x|/4>1\), which is \(|x|>4\). Therefore \(R=4\).
For a check, at \(x=1\) the limiting ratio is \(1/4<1\), so the series converges absolutely. At \(x=5\) it is \(5/4>1\), so the series diverges. Again, the results match the radius.
What the Radius Does Not Tell You
The radius separates points strictly inside the interval \(|x-c|<R\), where the power series converges absolutely, from points strictly outside it, where the series diverges. It does not by itself decide convergence at the boundary \(|x-c|=R\). At such a point, the geometric part of the term has absolute value \(1\), so the polynomial factor may still matter to the boundary series. Each boundary point requires its own test.
A common mistake is to assume that multiplying by a polynomial changes the radius because it makes coefficients larger. For example, seeing the factor \(n^2\) in \((n+1)^2 2^n\) might lead to the incorrect conclusion that its radius is less than \(1/2\). The root-test calculation corrects this: \(\bigl((n+1)^2\bigr)^{1/n}\to1\), while \((2^n)^{1/n}=2\). Thus the coefficient-root limit remains \(2\), giving \(R=1/2\). The polynomial can affect behavior at the boundary, but it does not shift the radius.
Summary
For \(q>0\), coefficients \(q^n\) give radius \(1/q\), while coefficients \(q^{-n}\) give radius \(q\). Multiplying either sequence by a nonzero polynomial in \(n\) leaves that radius unchanged, because the \(n\)th root of a polynomial factor tends to \(1\). These calculations identify convergence strictly inside and divergence strictly outside the radius; they do not settle the boundary points.
Check Your Understanding
Use the geometric ratio or the root-test radius formula to answer each question. Do not infer boundary behavior from the radius alone.
- What is the radius for a power series centered at \(0\) with coefficients \((2/7)^n\)?
- What is the radius for coefficients \(5^{-n}\)?
- Find the radius when the coefficients are \((n+1)^4 3^n\).
- Find the radius when the coefficients are \(n^2 6^{-n}\), with the initial zero coefficient understood to have no effect.
- Why does multiplying geometric coefficients by a nonzero polynomial in \(n\) leave the radius unchanged?
- Does knowing the radius alone determine whether the series converges at a point exactly that distance from its center?