Term-by-Term Differentiation of Function Series

Why Differentiating a Function Series Needs Control

For a finite sum, differentiation is linear: the derivative of a sum is the sum of the derivatives. For an infinite series, the corresponding statement is not automatic. If the partial sums converge to a function $S$, it does not follow merely from that convergence that their derivatives converge to $S'$, or even that $S$ is differentiable. The derivative series needs its own convergence control.

The useful condition is uniform convergence of the derivative series. One more condition is needed: the original series must converge at at least one point. Together, these hypotheses control the partial sums everywhere on a closed, bounded interval. This is the series version of the Local Uniform Derivative Criterion from “When Differentiation and Limits Commute,” and it gives a practical way to differentiate certain power series and other function series.

Definition: Let $a<b$, and suppose each $f_n:[a,b]\to\mathbb{R}$ is continuous on $[a,b]$ and differentiable on $(a,b)$. If both $\sum_{n=1}^{\infty}f_n(x)$ and $\sum_{n=1}^{\infty}f_n'(x)$ converge, term-by-term differentiation means that the sum $S$ of the first series is differentiable on $(a,b)$ and $$ S'(x)=\sum_{n=1}^{\infty}f_n'(x) \qquad (a<x<b). $$ The theorem below gives sufficient conditions for this conclusion.

The Term-by-Term Differentiation Theorem

Theorem (Term-by-Term Differentiation on a Closed Interval): Let $a<b$, and suppose each $f_n:[a,b]\to\mathbb{R}$ is continuous on $[a,b]$ and differentiable on $(a,b)$, with $f_n'$ continuous on $[a,b]$. Suppose the derivative series $\sum_{n=1}^{\infty}f_n'$ converges uniformly on $[a,b]$, and suppose $\sum_{n=1}^{\infty}f_n(x_0)$ converges for some $x_0\in[a,b]$. Then $\sum_{n=1}^{\infty}f_n$ converges uniformly on $[a,b]$ to a continuous function $S$. The function $S$ is differentiable on $(a,b)$, and $$ S'(x)=\sum_{n=1}^{\infty}f_n'(x) \qquad (a<x<b). $$

Proof. Let $$ P_N(x)=\sum_{n=1}^{N}f_n(x), \qquad Q_N(x)=\sum_{n=1}^{N}f_n'(x). $$ By hypothesis, $Q_N$ converges uniformly on $[a,b]$, so it is uniformly Cauchy. Also, the numerical sequence $P_N(x_0)$ converges, so it is Cauchy.

For integers $m>n$, the function $P_m-P_n$ is continuous on $[a,b]$ and differentiable on $(a,b)$. By the Mean Value Theorem, for each $x\in[a,b]$, $$ |(P_m-P_n)(x)| \leq |(P_m-P_n)(x_0)|+(b-a)\sup_{t\in[a,b]}|(Q_m-Q_n)(t)|. $$ Indeed, the difference between the values at $x$ and $x_0$ is at most $|x-x_0|$ times a bound for the derivative, and $|x-x_0|\leq b-a$. Both terms on the right tend to zero as $m,n\to\infty$. Thus $(P_N)$ is uniformly Cauchy. The Uniform Cauchy Criterion gives uniform convergence on $[a,b]$ to a function $S$. Since each $P_N$ is continuous, the theorem Uniform Limits of Continuous Functions Are Continuous shows that $S$ is continuous.

Let $g$ be the uniform limit of $Q_N$. Each $Q_N$ is continuous, so $g$ is continuous by the same uniform-limit theorem. For every $N$ and $x\in[a,b]$, the Fundamental Theorem of Calculus gives $$ P_N(x)-P_N(x_0)=\int_{x_0}^{x}Q_N(t)\,dt. $$ The Term-by-Term Integration Theorem applies to the uniformly convergent series $\sum f_n'$. Passing to the limit in this identity gives $$ S(x)-S(x_0)=\int_{x_0}^{x}g(t)\,dt. $$ Because $g$ is continuous, the Fundamental Theorem of Calculus implies $S'(x)=g(x)$ for every $x\in(a,b)$. By definition, $g(x)=\sum_{n=1}^{\infty}f_n'(x)$. This proves the assertion. $\square$

The proof uses the two hypotheses for different tasks. Convergence at $x_0$ anchors the values of the partial sums. Uniform convergence of the derivative series controls how much those partial sums can change as $x$ moves away from $x_0$. The closed, bounded interval supplies the finite distance factor $b-a$; the same argument does not give uniform convergence on an unbounded interval.

Estimating the Differentiation Remainder

The theorem also gives an error bound for replacing the full sum by its first $N$ terms. Write $g=\sum_{n=1}^{\infty}f_n'$, and let $P_N=\sum_{n=1}^{N}f_n$. The error at the anchor point is controlled by the numerical series there; the change in error across the interval is controlled by the tail of the derivative series.

Theorem (Derivative-Tail Remainder Estimate): Under the hypotheses of the Term-by-Term Differentiation Theorem, for every $N\geq1$, $$ \sup_{x\in[a,b]}|S(x)-P_N(x)| \leq |S(x_0)-P_N(x_0)| +(b-a)\sup_{x\in[a,b]}|g(x)-P_N'(x)|. $$

Proof. The proof of the differentiation theorem gives $$ S(x)-S(x_0)=\int_{x_0}^{x}g(t)\,dt. $$ For the finite sum $P_N$, the Fundamental Theorem of Calculus gives $P_N(x)-P_N(x_0)=\int_{x_0}^{x}P_N'(t)\,dt$. Subtract these identities. For every $x\in[a,b]$, $$ |S(x)-P_N(x)| \leq |S(x_0)-P_N(x_0)| +\left|\int_{x_0}^{x}(g(t)-P_N'(t))\,dt\right|. $$ The last term is at most $|x-x_0|\sup_{t\in[a,b]}|g(t)-P_N'(t)|$, and $|x-x_0|\leq b-a$. Taking the supremum over $x$ proves the estimate. $\square$

Both terms on the right tend to zero: the first because the original series converges at $x_0$, and the second because the derivative series converges uniformly. The estimate is useful when a quantitative error bound is needed, not just a proof that the partial sums converge.

Worked Applications

Worked Example: Differentiating the Logarithm Series

Fix $r$ with $0<r<1$, and consider on $[-r,r]$ the series $$ \sum_{n=1}^{\infty}(-1)^{n+1}\frac{x^n}{n}. $$ For $f_n(x)=(-1)^{n+1}x^n/n$, its derivative is $f_n'(x)=(-1)^{n+1}x^{n-1}$. For every $x\in[-r,r]$, $$ |f_n'(x)|=|x|^{n-1}\leq r^{n-1}. $$ The numerical geometric series $\sum_{n=1}^{\infty}r^{n-1}$ converges, so the Weierstrass M-Test gives uniform convergence of the derivative series. At $x_0=0$, every $f_n(0)=0$, so the original series converges there. The differentiation theorem applies.

The derivative series is geometric. Its $N$th partial sum is $$ \sum_{n=1}^{N}(-1)^{n+1}x^{n-1} =\sum_{k=0}^{N-1}(-x)^k =\frac{1-(-x)^N}{1+x}. $$ Since $|x|^N\leq r^N\to0$ uniformly on $[-r,r]$, its sum is $1/(1+x)$. The Logarithm Series identifies the original sum as $\ln(1+x)$. Therefore, term-by-term differentiation gives $$ \frac{d}{dx}\ln(1+x) =\sum_{n=1}^{\infty}(-1)^{n+1}x^{n-1} =\frac{1}{1+x} \qquad (|x|<1). $$ The argument applies on every interval $[-r,r]$ with $r<1$, and hence establishes the identity throughout $(-1,1)$.

Worked Example: Differentiating the Sine Series

Use the sine and cosine series $S(x)$ and $C(x)$ introduced in “The Cosine Series.” On a fixed interval $[-R,R]$, where $R>0$, write the sine series as $$ S(x)=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!}. $$ The derivative of its $n$th term is $(-1)^n x^{2n}/(2n)!$. For $|x|\leq R$, its absolute value is at most $R^{2n}/(2n)!$. The numerical series $\sum_{n=0}^{\infty}R^{2n}/(2n)!$ converges, so the derivative series converges uniformly by the Weierstrass M-Test. The original series converges at $x_0=0$, where all its terms vanish.

The theorem therefore permits differentiation term by term on $[-R,R]$. The resulting series is exactly the cosine series: $$ S'(x)=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!}=C(x) \qquad (-R<x<R). $$ Because $R$ can be chosen larger than any specified $|x|$, this identity holds for every real $x$. The factorial majorant is what provides uniform control on each bounded interval.

Worked Example: Differentiating a Binomial Series

Consider the binomial series with exponent $1/2$: $$ \sum_{n=0}^{\infty}\binom{1/2}{n}x^n. $$ Fix $0<r<1$ and work on $[-r,r]$. The $n$th term has derivative $n\binom{1/2}{n}x^{n-1}$ when $n\geq1$. The coefficients satisfy $|\binom{1/2}{n}|\leq1$: the first coefficient has absolute value $1/2$, and for $n\geq1$ the ratio of consecutive absolute values is $$ \frac{\left|\binom{1/2}{n+1}\right|}{\left|\binom{1/2}{n}\right|} =\frac{n-1/2}{n+1}<1. $$ Thus, for $|x|\leq r$, $$ \left|n\binom{1/2}{n}x^{n-1}\right| \leq nr^{n-1}. $$ The series $\sum_{n=1}^{\infty}nr^{n-1}$ converges, so the derivative series converges uniformly by the Weierstrass M-Test. At $x_0=0$, the original series has value $1$, and therefore converges.

The Term-by-Term Differentiation Theorem applies. By the Binomial Series, the original series sums to $(1+x)^{1/2}$ for $|x|<1$. Consequently, $$ \frac{d}{dx}(1+x)^{1/2} =\sum_{n=1}^{\infty}n\binom{1/2}{n}x^{n-1} =\frac{1}{2\sqrt{1+x}} \qquad (|x|<1). $$ For the last equality, differentiate the function $(1+x)^{1/2}$ using the ordinary power rule. The uniform majorant justifies identifying its derivative with the displayed series.

Why Pointwise Convergence of Derivatives Is Not Enough

Uniform convergence of the derivative series cannot be replaced by pointwise convergence alone. The following example shows that even when the partial sums converge pointwise and their derivatives converge pointwise, the limit function may fail to be differentiable.

Worked Example: Pointwise Derivative Convergence with a Nondifferentiable Sum

On $[-1,1]$, define $G_0(x)=0$ and, for $n\geq1$, set $$ G_n(x)=\sqrt{x^2+\frac{1}{n}}. $$ Each $G_n$ is continuously differentiable, with $$ G_n'(x)=\frac{x}{\sqrt{x^2+1/n}}. $$ Define $f_n=G_n-G_{n-1}$ for $n\geq1$. Then the partial sums telescope: $$ \sum_{k=1}^{N}f_k(x)=G_N(x)-G_0(x)=\sqrt{x^2+\frac{1}{N}}. $$ For each fixed $x$, these partial sums converge to $|x|$. Thus the function series converges pointwise to $S(x)=|x|$.

The derivatives of the partial sums are $G_N'(x)$. If $x>0$, then $$ \lim_{N\to\infty}G_N'(x) =\lim_{N\to\infty}\frac{x}{\sqrt{x^2+1/N}}=1. $$ If $x<0$, the same calculation gives the limit $-1$. At $x=0$, every $G_N'(0)=0$. Hence the derivative series, whose partial sums are $G_N'$, converges pointwise to a function that equals $1$ for $x>0$, $-1$ for $x<0$, and $0$ at $0$. However, $S(x)=|x|$ is not differentiable at $0$. The derivative convergence is not uniform: the derivatives change increasingly rapidly near zero as $N$ grows. This does not contradict the theorem, whose uniform-convergence hypothesis fails.

A second common pitfall is to omit convergence of the original series at the anchor point. For instance, the constant terms $f_n(x)=1$ have derivative series $\sum f_n'(x)=0$, which converges uniformly, but $\sum f_n(x)$ diverges at every point. Uniform control of derivatives alone cannot determine the values of the sum; the anchor condition supplies that missing information.

Choose a closed, bounded interval.
For power series, work on a smaller interval strictly inside the radius of convergence.

Control the derivative series uniformly.
Use the Weierstrass M-Test or another uniform tail estimate.

Check one anchor value.
Verify convergence of the original series at some point of the interval.

Differentiate the sum.
Apply the theorem to conclude that the sum is differentiable in the interior and its derivative is the derivative-series sum.

Takeaway: On a closed, bounded interval, uniform convergence of the derivative series together with convergence of the original series at one point justifies term-by-term differentiation in the interior. Pointwise convergence of derivatives, without uniform control, is not enough.

Check Your Understanding

Use the theorem, its proof, and the examples above to answer the following questions.

What two convergence hypotheses accompany the differentiability assumptions in the Term-by-Term Differentiation Theorem?
Where does the factor $b-a$ enter the proof that the partial sums converge uniformly?
What two quantities control the remainder estimate for the partial sum $P_N$?
Why does the binomial-series example use an interval $[-r,r]$ with $r<1$?
In the example with $G_N(x)=\sqrt{x^2+1/N}$, what fails in the hypotheses of the theorem?