Why the Radius Does Not Change
A power series can converge on a whole interval around its center, and its radius records how far that interval extends. Differentiating each term changes the coefficients: the coefficient of the \(n\)th power is multiplied by an index. At first, this may seem capable of changing the convergence range substantially.
It does not change the radius. The reason is that multiplying coefficients by an index changes their size only by a factor whose \(n\)th root tends to \(1\). The radius depends on the coefficients' exponential growth, so this extra factor leaves it unchanged.
Power Series and Radius
Consider a power series centered at \(c\):
Its radius of convergence \(R\) is the number such that the series converges absolutely whenever \(|x-c|<R\) and diverges whenever \(|x-c|>R\). The radius may be \(0\) or \(+\infty\). At the boundary \(|x-c|=R\), when \(0<R<\infty\), either convergence or divergence is possible and must be investigated separately.
The root-test formula, also called the Cauchy–Hadamard formula, expresses the radius in terms of the coefficients:
Here the conventions are \(1/0=+\infty\) and \(1/(+\infty)=0\). A finite number of initial coefficients does not affect the radius.
Termwise differentiation formally gives:
The last expression is again a power series centered at \(c\), with coefficients \(b_k=(k+1)a_{k+1}\). The radius question is therefore whether replacing \(a_k\) with \(b_k\) changes the root-test limit superior.
Proof That the Radii Are Equal
Write \(L=\limsup_{n\to\infty}|a_n|^{1/n}\), so \(R=1/L\) with the conventions above. For \(k\geq 1\), the \(k\)th root of the absolute value of the differentiated coefficient is
The factor \((k+1)^{1/k}\) tends to \(1\), and the exponent \((k+1)/k\) tends to \(1\). These facts ensure the limit superior of the displayed expression is \(L\). Here is why, including the extreme cases.
Set \(q_n=|a_n|^{1/n}\). For any \(\varepsilon>0\), eventually \(q_n\leq L+\varepsilon\), so \(q_{k+1}^{(k+1)/k}\) has limit superior at most \(L+\varepsilon\). Along a subsequence for which \(q_{k+1}\) tends to \(L\), the same expression tends to \(L\). Letting \(\varepsilon\) decrease to zero gives limit superior \(L\).
Then \(q_{k+1}\) tends to zero. Eventually \(q_{k+1}<1\), and since \((k+1)/k>1\), we have \(q_{k+1}^{(k+1)/k}\leq q_{k+1}\). Thus the transformed roots tend to zero.
There is a subsequence along which \(q_{k+1}\) tends to \(+\infty\). Raising those terms to the exponent \((k+1)/k>1\) still gives a sequence tending to \(+\infty\). The transformed roots therefore have infinite limit superior.
In all cases, multiplying by \((k+1)^{1/k}\), which tends to \(1\), leaves the limit superior unchanged. Therefore
The root-test formula now gives the same radius \(1/L=R\) for both series. This proves the theorem. The termwise differentiation theorem also says that, for every \(x\) satisfying \(|x-c|<R\), the sum of the derivative series equals the derivative of the original sum. That identity is an interior statement, not a claim about boundary points.
Worked Examples
Worked Example: A Geometric Series Centered Away from Zero
Consider
This is geometric with ratio \((x-2)/3\). It converges when \(|x-2|<3\), so its radius is \(3\). Within that interval, its sum is
Differentiate each term and reindex using \(k=n-1\):
The \(k\)th coefficient has \(k\)th root \(\left((k+1)/3^{k+1}\right)^{1/k}\), which tends to \(1/3\). Thus the differentiated series has radius \(1/(1/3)=3\). As a check of the differentiated sum, differentiating the closed form gives \(f'(x)=3/(5-x)^2\). At the center \(x=2\), the differentiated series has value \(1/3\), and the formula gives \(3/(5-2)^2=1/3\).
Worked Example: Equal Radii, Different Boundary Behavior
Consider the series
The coefficient roots are \((1/n^2)^{1/n}\), which tend to \(1\). Hence the radius is \(1\). Termwise differentiation for \(|x|<1\) gives
The differentiated coefficients \(1/(k+1)\) have \(k\)th roots tending to \(1\), so this series also has radius \(1\).
Now inspect the boundary. At \(x=1\), the original series is \(\sum_{n=1}^{\infty}1/n^2\), which converges, but the derivative series is \(\sum_{k=0}^{\infty}1/(k+1)\), the divergent harmonic series. At \(x=-1\), the original series converges absolutely, and the derivative series becomes \(\sum_{k=0}^{\infty}(-1)^k/(k+1)\), which converges by the alternating series test. The radii are equal even though these endpoint verdicts differ. The interior theorem alone does not assert that the derivative series converges at \(x=1\), nor does its interior identity establish a derivative there.
Worked Example: An Infinite Radius Remains Infinite
Consider the exponential power series
For \(n\geq 2\), at least \(n/2\) of the factors in \(n!\) are at least \(n/2\), so \(n!\geq(n/2)^{n/2}\). Consequently, \((1/n!)^{1/n}\leq(2/n)^{1/2}\), which tends to zero. The root-test formula gives infinite radius.
Differentiating term by term and setting \(k=n-1\) gives
The differentiated series is the original series, so it also has infinite radius. At \(x=0\), both series have value \(1\), which checks the constant term of the differentiated series directly.
What the Theorem Does Not Say
For a positive finite radius \(R\), the theorem identifies convergence inside the interval \((c-R,c+R)\) and divergence outside its closed interval. It does not supply a verdict at \(c-R\) or \(c+R\). The boundary example above makes this limitation concrete: a series may converge at an endpoint while its derivative series diverges there.
Nor does equality of radii mean that the two series have the same coefficients or the same sums. Differentiation changes the coefficients, and the sums are related by differentiation only where the termwise differentiation theorem applies: strictly inside the radius. Boundary convergence, endpoint derivatives, and endpoint identities require separate reasoning.
The practical lesson is to separate two questions. First, use the coefficient growth to find the radius; the derivative's index factor does not change it. Second, when an endpoint matters, test the original or differentiated series at that point directly. The radius theorem answers the first question, not the second.
Check Your Understanding
Use the coefficient root test and the distinction between interior and boundary behavior in your answers.
- Write the coefficient of \((x-c)^k\) in the termwise derivative of \(\sum_{n=0}^{\infty}a_n(x-c)^n\).
- Why does multiplying a coefficient by an index leave the root-test limit superior unchanged?
- If the original series has radius \(R=+\infty\), what is the radius of its derivative series?
- What does the termwise differentiation theorem guarantee at points satisfying \(|x-c|<R\)?
- Does equality of radii determine convergence of the derivative series at either endpoint? Explain briefly.
