Why Integration Does Not Change the Radius
A power series can converge on a particular interval around its center, and termwise integration changes every coefficient after the constant term. The new coefficients are divided by positive integers. This may make them smaller, but it does not change their exponential rate of growth, which is what determines the radius of convergence.
The integration constant has a different role. It supplies the constant term of the antiderivative; it is not generated by integrating any of the positive powers in the original series. Since changing one coefficient cannot change the radius, the constant does not affect the result.
The Series and Its Antiderivative
Let \(a\) be the center of a power series \[ f(x)=\sum_{n=0}^{\infty}c_n(x-a)^n \] with radius of convergence \(R\). Within the interval \(|x-a|<R\), the termwise integration theorem gives an antiderivative in the form \[ F(x)=C+\sum_{n=0}^{\infty}\frac{c_n}{n+1}(x-a)^{n+1}, \] where \(C\) is an arbitrary constant. The series part has no constant term: each of its powers has exponent at least \(1\). In a power series centered at \(a\), the number \(C\) is precisely the coefficient of \((x-a)^0\).
For example, differentiating the displayed expression term by term gives \[ F'(x)=\sum_{n=0}^{\infty}\frac{c_n}{n+1}(n+1)(x-a)^n =\sum_{n=0}^{\infty}c_n(x-a)^n=f(x) \] for \(|x-a|<R\). This also shows why the denominator \(n+1\) appears: it cancels the factor produced by differentiating the power \((x-a)^{n+1}\).
Radius of Convergence and Coefficient Growth
The root-test formula describes the radius of a power series directly from its coefficients. For a series \(\sum_{n=0}^{\infty}c_n(x-a)^n\), put \[ \alpha=\limsup_{n\to\infty}|c_n|^{1/n}. \] Then its radius is \(R=1/\alpha\), with the conventions that \(1/0=\infty\) and \(1/\infty=0\). This formula determines convergence for \(|x-a|<R\) and divergence for \(|x-a|>R\); it does not by itself determine what happens at the boundary \(|x-a|=R\).
For the integrated series, the coefficient of \((x-a)^m\), for \(m\geq 1\), is \[ d_m=\frac{c_{m-1}}{m}, \] and its constant coefficient is \(d_0=C\). The radius depends on the limiting upper root growth of the coefficients. The constant \(d_0\) is just one coefficient, so it has no effect on that limiting behavior. The remaining question is whether shifting the index and dividing by \(m\) changes the limiting upper root growth. It does not.
Why the Coefficient Growth Is Unchanged
Here is a complete justification of the key limiting step. Set \(q_n=|c_n|^{1/n}\) for \(n\geq 1\). The \(n\)th root of the coefficient of \((x-a)^{n+1}\) in the integrated series is \[ \left|\frac{c_n}{n+1}\right|^{1/(n+1)} =q_n^{\,n/(n+1)}(n+1)^{-1/(n+1)}. \] The factor \((n+1)^{-1/(n+1)}\) tends to \(1\). Raising \(q_n\) to the exponent \(n/(n+1)\), which tends to \(1\), also leaves its limiting upper value unchanged. To make this last statement precise, consider the possible values of \(\alpha=\limsup q_n\).
Eventually the \(q_n\) are bounded above. On any bounded range of nonnegative values, replacing \(q_n\) by \(q_n^{n/(n+1)}\) changes the values by an amount tending to zero. Thus the transformed sequence has limiting upper value \(\alpha\).
For every \(\varepsilon>0\), eventually \(q_n<\varepsilon\). The transformed roots then have limiting upper value at most \(\varepsilon\), and hence have limiting upper value \(0\).
There is a subsequence along which \(q_n\) tends to infinity. Along that subsequence, \(q_n^{n/(n+1)}(n+1)^{-1/(n+1)}\) also tends to infinity, so the transformed sequence has limiting upper value infinity.
Consequently, the limiting upper root growth of the integrated coefficients is the same \(\alpha\) as that of the original coefficients. The coefficient at index \(0\) is finite and does not affect a limit as the index tends to infinity. Applying the root-test formula to both series therefore gives the same radius \(1/\alpha\), including the cases \(R=0\) and \(R=\infty\).
Worked Examples
Worked Example: Integrating a Geometric-Type Power Series
Consider \[ f(x)=\sum_{n=0}^{\infty}2^n(x-1)^n. \] The original coefficient roots are \(|2^n|^{1/n}=2\) for \(n\geq1\), so its radius is \(R=1/2\). Termwise integration gives \[ F(x)=5+\sum_{n=0}^{\infty}\frac{2^n}{n+1}(x-1)^{n+1}. \] For \(n\geq0\), the coefficient of \((x-1)^{n+1}\) is \(2^n/(n+1)\). Its \((n+1)\)th root is \[ \left(\frac{2^n}{n+1}\right)^{1/(n+1)} =\frac{2^{n/(n+1)}}{(n+1)^{1/(n+1)}}\longrightarrow 2. \] Thus the antiderivative series has radius \(1/2\), just like the original. To check the antiderivative, differentiate inside that radius: \[ F'(x)=\sum_{n=0}^{\infty}\frac{2^n}{n+1}(n+1)(x-1)^n =\sum_{n=0}^{\infty}2^n(x-1)^n=f(x). \] The term \(5\) differentiates to zero, as an integration constant should.
Worked Example: A Finite Radius with a Polynomial Factor
Let \[ f(x)=\sum_{n=0}^{\infty}\frac{n^2}{5^n}(x+2)^n. \] For \(n\geq1\), the coefficient roots are \[ \left(\frac{n^2}{5^n}\right)^{1/n} =\frac{n^{2/n}}{5}\longrightarrow \frac15, \] so \(R=5\). (The \(n=0\) coefficient is zero and does not affect the radius.) An antiderivative is \[ F(x)=-3+\sum_{n=0}^{\infty}\frac{n^2}{(n+1)5^n}(x+2)^{n+1}. \] For \(n\geq1\), the root of the coefficient of \((x+2)^{n+1}\) is \[ \left(\frac{n^2}{(n+1)5^n}\right)^{1/(n+1)} =\frac{n^{2/(n+1)}}{(n+1)^{1/(n+1)}5^{n/(n+1)}}\longrightarrow\frac15. \] The integrated series therefore also has radius \(5\). Differentiating verifies the result: \[ F'(x)=\sum_{n=0}^{\infty}\frac{n^2}{(n+1)5^n}(n+1)(x+2)^n =\sum_{n=0}^{\infty}\frac{n^2}{5^n}(x+2)^n=f(x). \] At \(n=0\), both corresponding terms are zero, so the calculation also covers the initial index.
Worked Example: An Infinite Radius of Convergence
Consider \[ f(x)=\sum_{n=0}^{\infty}\frac{(x-3)^n}{n!}. \] For \(n\geq1\), the coefficient roots are \((1/n!)^{1/n}\), which tend to \(0\). Hence \(R=\infty\). Termwise integration gives \[ F(x)=2+\sum_{n=0}^{\infty}\frac{(x-3)^{n+1}}{(n+1)n!}. \] The \((n+1)\)th root of each integrated coefficient is \[ \left(\frac{1}{(n+1)n!}\right)^{1/(n+1)}\longrightarrow 0, \] since \(n!\) grows faster than any fixed exponential. The integrated series also has infinite radius. Its derivative is \[ F'(x)=\sum_{n=0}^{\infty}\frac{n+1}{(n+1)n!}(x-3)^n =\sum_{n=0}^{\infty}\frac{(x-3)^n}{n!}=f(x). \]
The Constant Term and a Common Mistake
The constant of integration belongs in the coefficient of \((x-a)^0\), not among the coefficients obtained by integrating the original terms. In the general expression \[ F(x)=C+\sum_{n=0}^{\infty}\frac{c_n}{n+1}(x-a)^{n+1}, \] the series begins with a multiple of \((x-a)^1\), so its value at \(x=a\) is zero. Therefore \(F(a)=C\).
A common mistake is to say that the termwise integral without \(C\) is the most general antiderivative. Take the simplest possible series, \(f(x)=1\), centered at \(a=0\). Its termwise integral is \(x\), but the most general antiderivative is \(x+C\). Indeed, differentiating \(x\) gives \(1\), and differentiating \(x+C\) also gives \(1\); the derivative cannot determine \(C\). The correction is to include an arbitrary constant term. Omitting it gives one antiderivative, not the full family.
Another tempting error is to infer that the integrated series must converge or diverge at every boundary point exactly as the original does. The theorem proves equality of the radii, not equality of boundary behavior. At points with \(|x-a|=R\), each series must be tested separately; the root-test formula for the radius does not settle those cases.
Why the Result Matters
The radius of convergence describes how far from the center the series represents a convergent power-series expression. The theorem shows that taking an antiderivative does not shrink or enlarge that radius. The coefficients change, but only by the factor \(1/(n+1)\), which has no effect on their limiting upper root growth. This makes it possible to integrate power series without recalculating a new radius from scratch.
The constant is equally important for interpreting the result correctly. The coefficient operation determines the changing-power part of an antiderivative, while the constant coefficient records the one value that differentiation loses. Keeping those roles separate makes both the series formula and its radius transparent.
Check Your Understanding
Answer each question using the coefficient formula and the root-test description of radius.
- Write the termwise antiderivative of \(\sum_{n=0}^{\infty}c_n(x-a)^n\), including its arbitrary constant.
- Why does division of the \(n\)th coefficient by \(n+1\) leave the limiting upper root growth unchanged?
- If the original power series has radius \(R=4\), what is the radius of its termwise antiderivative series, and why?
- Where does the integration constant appear when the antiderivative is written as a power series centered at \(a\)?
- Does equality of radii imply that the original and integrated series have the same convergence behavior at boundary points? Explain.