Term-by-Term Differentiation

From Continuity to Differentiation

The previous tutorial showed that a power series is continuous at every point strictly inside its radius of convergence. Differentiation asks for more: can we differentiate the sum by differentiating each term? For a finite sum of polynomials, the answer is immediate. For an infinite series, it requires control of the limit and the derivatives together.

Consider a power series centered at $a$, with partial sums $S_N$. Differentiating each finite partial sum gives another polynomial:

S_N(x)=\sum_{n=0}^{N}c_n(x-a)^n, \qquad S_N'(x)=\sum_{n=1}^{N}n c_n(x-a)^{n-1}.

This suggests the derivative series. The important result is that, strictly inside the original radius, this series converges and its sum is the derivative of the original power series sum.

Theorem (Term-by-Term Differentiation of a Power Series): Suppose $\sum_{n=0}^{\infty}c_n(x-a)^n$ has radius of convergence $R>0$, and let $f$ be its sum on $(a-R,a+R)$. Then $f$ is differentiable at every $x$ with $|x-a|<R$, and $$ f'(x)=\sum_{n=1}^{\infty}n c_n(x-a)^{n-1}. $$ The differentiated power series has radius of convergence $R$ as well.

The formula differentiates the term $c_n(x-a)^n$ to $n c_n(x-a)^{n-1}$; the constant term contributes zero. The theorem concerns points strictly inside the radius. It makes no general claim about differentiability at either endpoint. The proof that the limit function has the displayed derivative is the central issue in this topic. Before addressing that proof, we establish two properties of the differentiated series that will be useful: it converges uniformly on every smaller closed interval, and its radius is unchanged.

Uniform Convergence of the Differentiated Series

Uniform convergence inside the radius was essential for continuity. For differentiation, we also need uniform control of the series formed by the derivatives of the terms. The following result supplies that control on any closed interval whose distance from the center is strictly less than $R$.

Theorem (Uniform Convergence of the Derivative Series Inside the Radius): Let $\sum_{n=0}^{\infty}c_n(x-a)^n$ have radius of convergence $R>0$. For every finite $r$ with $0\leq r<R$, the series $$ \sum_{n=1}^{\infty}n c_n(x-a)^{n-1} $$ converges uniformly on $[a-r,a+r]$.

Proof. Choose a finite $s$ such that $r<s<R$. When $R=\infty$, choose any finite $s>r$. The Convergence Inside and Outside the Radius Theorem gives absolute convergence at distance $s$, so

\sum_{n=1}^{\infty}|c_n|s^n<\infty.

For $|x-a|\leq r$, the absolute value of the $n$th term of the derivative series is bounded by

n|c_n||x-a|^{n-1} \leq n|c_n|r^{n-1} =\frac{n}{s}\left(\frac{r}{s}\right)^{n-1}|c_n|s^n.

Set $q=r/s$, so $0\leq q<1$. The sequence $nq^{n-1}$ is bounded. For $q=0$, its first term is $1$ and all later terms are zero. For $0<q<1$, the ratio of consecutive terms is

\frac{(n+1)q^n}{nq^{n-1}} =q\left(1+\frac{1}{n}\right),

which is less than some fixed number strictly below $1$ for all sufficiently large $n$. Thus the sequence is eventually decreasing and is bounded by the maximum of its finitely many earlier terms and its first term. Let $M$ be a bound for $nq^{n-1}/s$. Then, for every $x\in[a-r,a+r]$,

n|c_n||x-a|^{n-1}\leq M|c_n|s^n.

The series of bounds $\sum_{n=1}^{\infty}M|c_n|s^n$ converges. Its tails therefore tend to zero, independently of $x$ in the interval. The tails of the derivative series are bounded by those same tails, which proves uniform convergence there. $\square$

This estimate explains why the strict inequality $r<R$ matters: it lets us choose the larger distance $s$ while staying inside the radius. The factor $n$ introduced by differentiation is controlled by the geometric factor $(r/s)^{n-1}$.

Why the Radius Does Not Change

Differentiation changes the coefficients and shifts the powers down by one. Despite those changes, it does not change the radius of convergence. We prove this directly, using absolute convergence inside a power series’ radius.

Theorem (The Differentiated Series Has the Same Radius): The power series $\sum_{n=1}^{\infty}n c_n(x-a)^{n-1}$ has the same radius of convergence as $\sum_{n=0}^{\infty}c_n(x-a)^n$.

Proof. Let the original radius be $R$, and let the radius of the derivative series be $Q$. If $0<r<R$, choose $s$ with $r<s<R$. The estimate in the proof of the Uniform Convergence of the Derivative Series Theorem shows that the derivative series converges absolutely whenever $|x-a|\leq r$. Since this holds for every $r<R$, its radius satisfies $Q\geq R$.

For the reverse inequality, take any $r$ with $0<r<Q$. At a point whose distance from $a$ is $r$, absolute convergence of the derivative series means

\sum_{n=1}^{\infty} n|c_n|r^{n-1}<\infty.

For every $n\geq1$, $r/n\leq r$, so

|c_n|r^n =\frac{r}{n}n|c_n|r^{n-1} \leq r\,n|c_n|r^{n-1}.

Comparison therefore gives $\sum_{n=1}^{\infty}|c_n|r^n<\infty$; adding the single term $|c_0|$ does not affect convergence. Hence the original power series converges absolutely at distance $r$, so $R\geq r$. This is true for every $0<r<Q$, which gives $R\geq Q$. If $Q=0$, that inequality is automatic; if $Q=\infty$, the argument gives $R\geq r$ for every finite $r>0$, so $R=\infty$. In all cases $R=Q$. $\square$

The proof compares the two series in opposite directions. Inside the original radius, the extra factor $n$ is offset by choosing a slightly larger distance $s$. In the reverse direction, multiplying a derivative-series term by $r/n$ recovers the corresponding original term.

Worked Examples: Differentiating Power Series

Worked Example: Differentiating a Geometric Series

Consider the power series centered at $-2$:

\sum_{n=0}^{\infty}\frac{(x+2)^n}{4^n}.

It is geometric with ratio $t=(x+2)/4$, so its radius is $4$. For $|x+2|<4$, its sum is $1/(1-t)=4/(2-x)$. The Term-by-Term Differentiation Theorem gives

\frac{d}{dx}\sum_{n=0}^{\infty}\frac{(x+2)^n}{4^n} =\sum_{n=1}^{\infty}\frac{n(x+2)^{n-1}}{4^n}.

Differentiating the explicit sum gives the same result as a function:

\frac{d}{dx}\left(\frac{4}{2-x}\right) =\frac{4}{(2-x)^2}.

Indeed, differentiating $4(2-x)^{-1}$ gives $4(2-x)^{-2}$, since the derivative of $2-x$ is $-1$. The series formula and the explicit derivative are valid throughout $-6<x<2$, not at the endpoints by this theorem.

Worked Example: A Factorial-Coefficient Series

Let

f(x)=\sum_{n=0}^{\infty}\frac{(-1)^n(x-1)^{2n+1}}{(2n+1)!}.

The ratio of the absolute values of consecutive nonzero terms at a fixed $x\ne1$ is

\frac{|x-1|^{2n+3}/(2n+3)!}{|x-1|^{2n+1}/(2n+1)!} =\frac{|x-1|^2}{(2n+3)(2n+2)} \longrightarrow 0.

The Ratio Test gives convergence for every real $x$, so the radius is infinite. Term-by-term differentiation is therefore valid for every $x$, and gives

f'(x)=\sum_{n=0}^{\infty}\frac{(-1)^n(x-1)^{2n}}{(2n)!}.

The index begins at $n=0$ in this display because differentiating the term with exponent $2n+1$ cancels its factor $2n+1$ in the factorial: $(2n+1)/(2n+1)!=1/(2n)!$. The first terms confirm the pattern: $f(x)=(x-1)-(x-1)^3/3!+(x-1)^5/5!-\cdots$, while $f'(x)=1-(x-1)^2/2!+(x-1)^4/4!-\cdots$.

Worked Example: A Convergent Endpoint Does Not Settle Differentiability

Consider

g(x)=\sum_{n=1}^{\infty}\frac{x^n}{n^2}.

The root test gives radius $1$, since $(1/n^2)^{1/n}\to1$. For $|x|<1$, term-by-term differentiation yields

g'(x)=\sum_{n=1}^{\infty}\frac{x^{n-1}}{n}.

At $x=1$, the original series is $\sum_{n=1}^{\infty}1/n^2$, which converges by the p-Series Convergence Criterion. But the displayed derivative series at $x=1$ would be $\sum_{n=1}^{\infty}1/n$, which diverges by the same criterion. This does not contradict the theorem: $x=1$ is an endpoint, not a point strictly inside the radius. Convergence of the original series at that endpoint alone does not allow the term-by-term differentiation formula to be applied there.

What the Theorem Does—and Does Not—Permit

The theorem gives a powerful conclusion at every interior point: the sum is differentiable, and its derivative is represented by the series obtained by differentiating each term. The uniform convergence result above is an important ingredient in the justification, but it is not by itself a proof that the derivative of the limit equals the limit of the derivatives. That step requires a separate argument linking the partial sums to the limit function.

The endpoint distinction is equally important. A power series may converge at an endpoint while its differentiated series diverges there, as the reciprocal-square example shows. The theorem guarantees differentiation on the open interval of convergence, not at its boundary. Endpoint differentiability must be investigated separately using the behavior of the sum function near the endpoint.

Finally, the differentiated series is itself a power series with the same radius $R$. This means it converges absolutely and uniformly on every closed interval strictly inside that radius, just as the original series does. It does not mean that the original and differentiated series have identical convergence behavior at the endpoints.

Takeaway: Inside its radius of convergence, a power series can be differentiated term by term. The resulting power series has the same radius, and converges uniformly on every smaller closed interval. Endpoint behavior is not covered by this conclusion.

Check Your Understanding

Use the definitions and results in this tutorial to answer the following questions.

What is the $n$th term of the derivative series obtained from $\sum_{n=0}^{\infty}c_n(x-a)^n$?
Why can a distance $s$ with $r<s<R$ control the factor $n$ in the derivative-series terms on $[a-r,a+r]$?
How does absolute convergence of the derivative series at distance $r>0$ imply convergence of the original series there?
What radius of convergence does the differentiated series have, and which theorem establishes this?
Why does convergence of an original power series at an endpoint not by itself justify differentiating there?

What You'll Learn