Why the Radius Is Stable
A power series can be differentiated or integrated term by term within its interval of convergence. Those operations change its coefficients: differentiation introduces a factor of \(n\), while integration divides by a factor involving \(n\). It is natural to wonder whether this change also alters the radius of convergence.
It does not. The key is that a power series’ radius is determined by the exponential rate at which its coefficients grow or shrink. Multiplying or dividing a coefficient by an index factor such as \(n\) does not change that exponential rate. But a radius describes the open interval around the center. At the two endpoints, even a small coefficient change can change whether the resulting series converges.
Radius and Coefficient Growth
For a power series centered at \(c\), the radius of convergence \(R\) measures how far from \(c\) the series converges absolutely. It converges absolutely when \(|x-c|<R\), and diverges when \(|x-c|>R\). At the boundary, where \(|x-c|=R\), the series may converge or diverge, so the radius alone does not settle the endpoint cases.
The root test makes the connection between the radius and coefficient growth precise. Let \(L=\limsup_{n\to\infty}|a_n|^{1/n}\). The radius is \(R=1/L\), using \(1/0=\infty\) and \(1/\infty=0\). Indeed, for a fixed \(x\), the \(n\)th root of the absolute value of the \(n\)th term is \(|a_n|^{1/n}|x-c|\). If \(L|x-c|<1\), the root test gives absolute convergence; if \(L|x-c|>1\), the terms fail to tend to zero along a subsequence, so the series diverges.
We now compare the coefficient growth rates before and after termwise calculus. Differentiation changes the coefficient \(a_n\) to \(n a_n\) and shifts the power down by one. Integration changes it to \(a_n/(n+1)\) and shifts the power up by one. The index factors do not affect the limiting \(n\)th-root growth rate: \(n^{1/n}\) and \((1/n)^{1/n}\) both approach \(1\). Shifting the index by one does not change the limsup either. Thus neither operation changes \(L\), and since \(R=1/L\), neither changes \(R\).
Derivation of the Radius Result
Write \(t=x-c\) to keep the powers centered at zero. The derivative series, with its powers indexed from zero, is
The new coefficient of \(t^m\) is \(b_m=(m+1)a_{m+1}\). Its root growth rate is the same as that of the original coefficients:
Here the factor \((m+1)^{1/m}\) approaches \(1\), and changing the index from \(m+1\) to \(n\) does not alter the limiting upper growth rate. This remains true when the rate is zero or infinite. Therefore the derivative series has radius \(1/L=R\).
An antiderivative with value \(0\) at the center has series
For \(m\geq 1\), its coefficient is \(a_{m-1}/m\). Taking \(m\)th roots introduces a factor \(m^{-1/m}\), which approaches \(1\); shifting the original index by one again preserves the limsup. The integrated series therefore has the same radius \(R\). Adding any constant to the antiderivative changes only its constant term, which cannot affect the radius.
Worked Example: Differentiating and Integrating a Geometric Series
Worked Example: Differentiating and Integrating a Geometric Series
For \(|x|<1\), the geometric series is \(\sum_{n=0}^{\infty}x^n=1/(1-x)\), so its radius is \(1\). Differentiate term by term:
The coefficient of \(x^{n-1}\) has acquired the factor \(n\). For \(|x|<1\), the derivative of \(1/(1-x)\) is \(1/(1-x)^2\), so
As a check, differentiating \(1/(1-x)\) gives \(1/(1-x)^2\), as required. The derivative series still has radius \(1\). At \(x=1\), its terms are \(n\), which do not approach zero. At \(x=-1\), they are \(n(-1)^{n-1}\), whose magnitudes also do not approach zero. It diverges at both endpoints.
Integrating the original series term by term from \(0\) to \(x\) gives
Differentiating \(-\ln(1-x)\) gives \(1/(1-x)\), which checks the antiderivative. Its series still has radius \(1\). At \(x=1\), the series is the harmonic series and diverges. At \(x=-1\), it is \(\sum_{n=0}^{\infty}(-1)^{n+1}/(n+1)\), the alternating harmonic series, which converges. The two endpoints behave differently even though the radius is unchanged.
Worked Example: Endpoint Behavior Changes After Calculus
Worked Example: Endpoint Behavior Changes After Calculus
Consider \(\sum_{n=1}^{\infty}x^n/n^2\). Since \((1/n^2)^{1/n}\) approaches \(1\), its radius is \(R=1\). At \(x=1\), the series is \(\sum 1/n^2\), which converges; at \(x=-1\), it converges absolutely because its term magnitudes are also \(1/n^2\).
Differentiating term by term gives
Its coefficient root growth is still \(1\), so its radius is \(1\). At \(x=1\), it becomes the divergent harmonic series \(\sum 1/n\). At \(x=-1\), it becomes \(\sum_{n=1}^{\infty}(-1)^{n-1}/n\), which converges by the alternating-series test: its terms decrease in magnitude to zero.
Integrating the original series from \(0\) to \(x\) gives
Its radius remains \(1\). At either endpoint, the magnitude of its \(n\)th term is \(1/[n^2(n+1)]\), which is at most \(1/n^3\). The series therefore converges absolutely at both endpoints by comparison with the convergent \(p\)-series \(\sum 1/n^3\).
For a check on the integration, differentiating each term \(x^{n+1}/[n^2(n+1)]\) gives \(x^n/n^2\), the corresponding original term. Thus termwise calculus preserves the radius here, while the endpoint tests give different results for the original, derivative, and integral series.
Worked Example: A Series Centered Away from Zero
Worked Example: A Series Centered Away from Zero
Consider the series centered at \(x=1\):
The coefficient magnitudes are \(1/(n2^n)\), whose \(n\)th roots approach \(1/2\). Its radius is \(R=2\), so its open interval of convergence is \(-1<x<3\). At \(x=3\), the series becomes the alternating harmonic series and converges. At \(x=-1\), its terms are \(-1/n\), so it diverges.
Differentiation gives
The factor \(n\) from differentiating \((x-1)^n\) cancels the \(n\) in the denominator. The remaining coefficient root growth is still \(1/2\), so the derivative series has radius \(2\). At \(x=3\), its terms are \((-1)^{n+1}/2\); at \(x=-1\), they are \(1/2\). Neither sequence of terms approaches zero, so the derivative series diverges at both endpoints.
Integrating term by term gives
The radius remains \(2\). At either endpoint, the magnitude of the \(n\)th term is \(2/[n(n+1)]\). Since \(2/[n(n+1)]\leq 2/n^2\), the integral series converges absolutely at both endpoints. Differentiating each integrated term yields \((-1)^{n+1}(x-1)^n/(n2^n)\), checking the termwise integration.
Common Pitfall: Assuming Endpoints Are Preserved
A common error is to conclude that an endpoint must keep its convergence behavior because the radius stays the same. For example, the series \(\sum_{n=1}^{\infty}x^n/n^2\) converges at \(x=1\). A mistaken argument might say: “The derivative has the same radius, so it also converges at \(x=1\).” But the derivative series at that endpoint is
which diverges. The radius identifies the boundary location; it does not decide whether the boundary points themselves belong to the interval of convergence. After differentiating or integrating, substitute each endpoint into the new series and test it directly.
Use the coefficient root growth or another suitable power-series test.
Differentiate or integrate the terms using the power rule.
Index factors do not change the coefficients’ exponential growth rate.
Substitute each boundary value into the new series and use an appropriate convergence test.
Check Your Understanding
Use the radius result and endpoint reasoning to answer these questions.
- Why does multiplying each coefficient \(a_n\) by \(n\) leave the radius unchanged?
- If a power series has radius \(3\), what is the radius of its termwise integral?
- For \(\sum_{n=1}^{\infty}x^n/n^2\), what series results from termwise differentiation, and does it converge at \(x=1\)?
- Does knowing a power series’ radius determine convergence at either endpoint? Explain.
- For a series centered at \(c\) with radius \(R\), which two values must be tested separately after termwise calculus?
