Examples of Radius of Convergence

Turning Coefficient Growth into a Radius

The previous tutorial analyzed what happens at the boundary of a power series’ convergence region. Before reaching that boundary, one must know where it is. The Cauchy–Hadamard Formula gives the radius from the exponential growth rate of the coefficients, but a formula becomes most useful when it can be recognized in concrete examples.

For a series centered at \(a\),

write \(L=\limsup_{n\to\infty}|c_n|^{1/n}\). The Cauchy–Hadamard Formula says that the radius is \(R=1/L\), with \(R=\infty\) when \(L=0\), and \(R=0\) when \(L=\infty\). The center \(a\) affects the location of the interval or region of convergence, but not this coefficient-growth calculation. In the examples below, the main task is to identify the exponential part of \(|c_n|\), or to compare consecutive coefficients.

A polynomial factor such as \(n^3\) grows without bound, but it grows more slowly than any fixed positive exponential factor. Thus it can affect the behavior at the boundary without changing the radius. By contrast, multiplying coefficients by \(\lambda^n\) changes their exponential growth and therefore changes the radius. The examples make these distinctions precise.

Polynomial Factors and Exponential Growth

Proof. Put \(v_n=u_n^{1/n}\), so \(v_n\to1\) and \(|d_n|^{1/n}=v_n|c_n|^{1/n}\). Fix \(0<\varepsilon<1\). For all sufficiently large \(n\), \(1-\varepsilon\leq v_n\leq1+\varepsilon\). Therefore

Taking limsups gives the corresponding inequalities between the limsups, with the usual extended nonnegative values allowed. If \(L=\limsup |c_n|^{1/n}\) is finite, these inequalities give \((1-\varepsilon)L\leq\limsup |d_n|^{1/n}\leq(1+\varepsilon)L\). Letting \(\varepsilon\) decrease to zero proves equality, including \(L=0\). If \(L=\infty\), the lower inequality shows that the limsup for \(d_n\) is at least \((1-\varepsilon)\) times arbitrarily large values, hence is also infinity. The Cauchy–Hadamard Formula now gives equal radii. \(\square\)

For each fixed nonnegative integer \(m\), \(u_n=n^m\) satisfies the hypothesis (ignoring the harmless initial value at \(n=0\) when \(m>0\)): \( (n^m)^{1/n}=\exp(m\log n/n)\to1\). Thus multiplying coefficients by a fixed power of \(n\) does not change their radius. The same reasoning applies to many other factors whose \(n\)th roots tend to one.

When a Lower-Growth Sequence Is Added

Sometimes coefficients are sums rather than products. Cancellation can matter in general, but it cannot erase a coefficient sequence whose exponential growth is strictly larger than that of the sequence being added to it.

Proof. For the upper bound, fix \(q>L\). By the definitions of the limsups, for all sufficiently large \(n\), \(|c_n|^{1/n}\leq q\) and \(|d_n|^{1/n}\leq q\). Hence

It follows that \(\limsup |c_n+d_n|^{1/n}\leq q\); letting \(q\) decrease to \(L\) gives an upper bound of \(L\).

For the lower bound, choose any \(s,t\) with \(M<t<s<L\). There are infinitely many indices \(n\) for which \(|c_n|^{1/n}>s\), since \(s<L\). For all sufficiently large \(n\), \(|d_n|^{1/n}\leq t\), since \(t>M\). Along the infinitely many indices satisfying both conditions, the reverse triangle inequality gives

The \(n\)th root of the final expression tends to \(s\), because \(0<t/s<1\). Thus the limsup for \(|c_n+d_n|^{1/n}\) is at least \(s\). Since \(s\) can be chosen arbitrarily close to \(L\) from below, that limsup is at least \(L\). Together with the upper bound, it equals \(L\). \(\square\)

In particular, if two power series have distinct finite positive radii, adding their coefficients term by term gives a series with the smaller radius. The series with the smaller radius has the larger coefficient growth rate, so the theorem rules out cancellation at that dominant exponential scale.

Worked Examples: Computing the Radius

Choosing a Calculation and Reading Its Limits

For a new example, first inspect the coefficient structure. A geometric factor \(\lambda^n\) often reveals the radius directly. A fixed polynomial multiplier can be set aside using the Subexponential Factors Preserve the Radius Theorem. For factorials and products, the consecutive ratio may simplify more quickly; the Ratio Criterion for the Radius of a Power Series then converts its limit into the radius. The Cauchy–Hadamard Formula remains the general method when neither pattern is convenient.

A common mistake is to confuse the ratio criterion for the radius with the Ratio Test at a boundary point. If the coefficients are eventually nonzero and \(|c_{n+1}|/|c_n|\to1\), the Ratio Criterion gives radius one. At a boundary point, however, the Ratio Test may be inconclusive because the ratio of consecutive terms can tend to one; the Cauchy–Hadamard Formula determines the radius, not endpoint convergence. Another mistake is to infer endpoint convergence from the radius. The first worked example has a radius of five, but that calculation alone says nothing about the two points at distance five from its center.