Proof of Term-by-Term Differentiation

Why Term-by-Term Differentiation Needs a Proof

For each finite partial sum of a power series, differentiating term by term is ordinary polynomial differentiation. The difficulty is that the power series sum is a limit of these partial sums: differentiating a sequence of functions and then taking its limit is not automatically the same as taking the limit first and then differentiating. We need a result that makes this passage valid.

The key idea is to control the derivatives uniformly on each smaller closed interval and to know that the partial sums converge at one point. The derivative control handles changes in function values across the interval; the value at the fixed point determines the otherwise undetermined additive constant. We first prove this general principle and then apply it to power series.

A Criterion for Differentiating a Limit

Theorem (Uniform Derivative Convergence Criterion): Let $I$ be an open interval, and let $f_N:I\to\mathbb{R}$ be continuously differentiable for each $N$. Suppose that $f_N'$ converges uniformly on every compact subinterval of $I$ to a function $g$, and that $f_N(x_0)$ converges for some fixed $x_0\in I$. Then $f_N$ converges uniformly on every compact subinterval of $I$ to a function $f$. The function $g$ is continuous on $I$, and $f$ is differentiable with $f'=g$.

Proof. First we show that the functions $f_N$ converge locally uniformly. Choose any compact subinterval $K\subset I$ containing $x_0$. For $M,N$, the Mean Value Theorem applied to $f_M-f_N$ between $x_0$ and $x\in K$ gives

|f_M(x)-f_N(x)| \leq |f_M(x_0)-f_N(x_0)| +|x-x_0|\sup_{y\in K}|f_M'(y)-f_N'(y)|.

The first term tends to zero as $M,N\to\infty$, since $f_N(x_0)$ converges. The second tends to zero uniformly for $x\in K$, since $f_N'$ converges uniformly on $K$, and $|x-x_0|$ is bounded there. Thus $(f_N)$ is uniformly Cauchy on $K$, so it converges uniformly on $K$. Any compact subinterval not containing $x_0$ can be included in a larger compact subinterval of $I$ that does contain $x_0$. Therefore $f_N$ converges uniformly on every compact subinterval of $I$ to a function $f$.

The function $g$ is continuous on every compact subinterval because it is the uniform limit there of the continuous functions $f_N'$, by the Uniform Limits of Continuous Functions Are Continuous Theorem. Hence $g$ is continuous throughout $I$.

It remains to prove that $f'=g$. Fix $x\in I$, and take a nonzero $h$ small enough that the segment joining $x$ to $x+h$ lies in a compact subinterval $K\subset I$. For each $N$, the Mean Value Theorem gives a point $\xi_N$ between $x$ and $x+h$ such that

\frac{f_N(x+h)-f_N(x)}{h}=f_N'(\xi_N).

Compare this quotient with $g(x)$. Since $\xi_N\in K$,

\left|\frac{f_N(x+h)-f_N(x)}{h}-g(x)\right| \leq \sup_{y\in K}|f_N'(y)-g(y)| +\sup_{\substack{y\text{ between }x\text{ and }x+h}}|g(y)-g(x)|.

For fixed $h$, let $N\to\infty$. The left-hand quotient converges to $(f(x+h)-f(x))/h$, and the first term on the right tends to zero by uniform convergence of the derivatives. Consequently,

\left|\frac{f(x+h)-f(x)}{h}-g(x)\right| \leq \sup_{\substack{y\text{ between }x\text{ and }x+h}}|g(y)-g(x)|.

As $h\to0$, the right-hand side tends to zero because $g$ is continuous at $x$. The difference quotient of $f$ therefore tends to $g(x)$, proving $f'(x)=g(x)$. Since $x$ was arbitrary, $f'=g$ throughout $I$. $\square$

The estimate in the proof shows exactly how the two hypotheses work together. Uniform convergence of the derivatives controls the variation of the approximating functions, while convergence at $x_0$ anchors their values. Without the anchor, derivative information alone cannot determine the limit function’s constant value.

Applying the Criterion to a Power Series

Let $\sum_{n=0}^{\infty}c_n(x-a)^n$ have radius of convergence $R>0$, and let $S_N$ denote its $N$th partial sum. Each $S_N$ is a polynomial, so it is continuously differentiable, with

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

On every compact subinterval strictly inside $(a-R,a+R)$, the derivatives converge uniformly to the sum of the derivative series. This is the Uniform Convergence of the Derivative Series Inside the Radius Theorem established in the previous tutorial. Also, for every $N$, $S_N(a)=c_0$, so the partial sums converge at the fixed point $a$. These are precisely the hypotheses needed for the criterion.

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 $$ f'(x)=\sum_{n=1}^{\infty}n c_n(x-a)^{n-1}. $$

Proof. Set $I=(a-R,a+R)$ when $R$ is finite, and $I=\mathbb{R}$ when $R=\infty$. For each compact subinterval $K\subset I$, the Uniform Convergence of the Derivative Series Inside the Radius Theorem gives uniform convergence of $S_N'$ on $K$ to the sum $g$ of the derivative series. At $x_0=a$, we have $S_N(a)=c_0$ for every $N$, so $S_N(a)$ converges. The Uniform Derivative Convergence Criterion now shows that $S_N$ converges locally uniformly to a differentiable function whose derivative is $g$. The pointwise limit of $S_N$ is the original power series sum $f$, so that function is $f$. Thus $f'=g$, which is the claimed formula. $\square$

This proof relies on uniform convergence on smaller intervals, not merely pointwise convergence of the derivative series. It does not need uniform convergence all the way to the radius boundary: around any interior point, a compact interval can be chosen that remains strictly inside the radius.

Worked Examples: Using the Proof

Worked Example: A Series with Harmonic Coefficients

Consider the power series centered at $2$:

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

The coefficient root limit is $\lim_{n\to\infty}(1/n)^{1/n}=1$, so the radius of convergence is $1$. On $1<x<3$, term-by-term differentiation gives

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

The factor $n$ from differentiating the $n$th power cancels the denominator $n$. The derivative series is geometric, with ratio $x-2$, so its sum is

\sum_{n=1}^{\infty}(x-2)^{n-1} =\frac{1}{1-(x-2)} =\frac{1}{3-x}, \qquad 1<x<3.

At the center $x=2$, the derivative series has first term $1$ and all later terms $0$, agreeing with $1/(3-2)=1$. The differentiation theorem justifies the derivative formula throughout the open interval, without requiring an explicit formula for the original sum.

Worked Example: A Series with Exponents in Steps of Three

Let

F(x)=\sum_{n=0}^{\infty}\frac{(x+1)^{3n+2}}{(3n+2)!}.

For a fixed $x\ne-1$, the ratio of consecutive absolute terms is

\frac{|x+1|^{3n+5}/(3n+5)!}{|x+1|^{3n+2}/(3n+2)!} =\frac{|x+1|^3}{(3n+5)(3n+4)(3n+3)} \longrightarrow 0.

The Ratio Test therefore gives convergence for every real $x$, so the radius is infinite. Differentiating the term with index $n$ gives

\frac{d}{dx}\left(\frac{(x+1)^{3n+2}}{(3n+2)!}\right) =\frac{(x+1)^{3n+1}}{(3n+1)!}.

Thus the theorem gives

F'(x)=\sum_{n=0}^{\infty}\frac{(x+1)^{3n+1}}{(3n+1)!}

for every real $x$. At $x=-1$, every term in the derivative series is zero, consistent with the fact that every term of $F$ has exponent at least $2$ and hence derivative zero at that point.

Worked Example: Growing Coefficients Inside a Finite Radius

Consider

G(x)=\sum_{n=0}^{\infty}(n+1)^2\left(\frac{x+1}{3}\right)^n.

The ratio of consecutive coefficient magnitudes is

\frac{(n+2)^2/3^{n+1}}{(n+1)^2/3^n} =\frac{1}{3}\left(\frac{n+2}{n+1}\right)^2 \longrightarrow \frac{1}{3}.

The Ratio Criterion for the Radius of a Power Series gives radius $3$. Inside $|x+1|<3$, differentiation yields

G'(x)=\sum_{n=1}^{\infty}\frac{n(n+1)^2}{3^n}(x+1)^{n-1}.

At the center $x=-1$, only the $n=1$ term of this derivative series is nonzero, giving $G'(-1)=4/3$. The proof applies even though the coefficients grow: on each smaller interval, the geometric distance from the radius boundary controls the extra factor $n$. No conclusion about differentiability at $x=-4$ or $x=2$ follows from the interior theorem.

What the Hypotheses Prevent

Uniform convergence of the derivatives is not enough on its own. For instance, the constant functions $f_N(x)=N$ all have derivative zero, so their derivatives converge uniformly, but the functions do not converge at any point. Even if some limit function exists, derivative information determines changes in values rather than the value at a particular point. The convergence assumption at $x_0$ supplies that missing information.

There is a second common mistake: concluding that the theorem applies at an endpoint because the original power series happens to converge there. The argument uses a compact interval around the point on which the derivative series converges uniformly. At an endpoint, such an interval may extend beyond the radius, where convergence is no longer guaranteed. Endpoint differentiability therefore requires a separate argument.

Takeaway: To justify differentiating a limit, control the derivatives uniformly on compact intervals and anchor the approximating functions at one point. For power series, the partial sums share the value $c_0$ at the center, and the derivative series converges uniformly on every smaller interval; together these facts prove term-by-term differentiation inside the radius.

Check Your Understanding

Use the criterion and proof above to answer the following questions.

Why does uniform convergence of the derivatives alone fail to guarantee convergence of the functions?
In the Uniform Derivative Convergence Criterion, how does the Mean Value Theorem help show that the functions are uniformly Cauchy on a compact interval?
Why do the partial sums of a power series automatically converge at the center?
Which result from the previous tutorial supplies uniform convergence of the derivative series on smaller intervals?
Why does convergence of a power series at an endpoint not, by itself, justify differentiating there?