Endpoint Convergence After Differentiation

What Differentiation Changes at the Boundary

A power series can converge throughout the interior of its interval and still behave differently at its endpoints. Differentiating its terms changes the coefficients, so it can also change what happens at those boundary points. The family \( \sum_{n=1}^{\infty} x^n/n^p \) makes that change especially clear: differentiation lowers the exponent on \(n\) by one.

We will compare the original series and the series obtained by differentiating each term. The comparison is at \(x=1\) and \(x=-1\), and it depends on the real parameter \(p\). The key qualification is that term-by-term differentiation is guaranteed inside the interval of convergence; endpoint behavior must be tested separately.

Definition: For a real number \(p\), let \( f_p(x)=\sum_{n=1}^{\infty}x^n/n^p \) where the series converges. Its termwise differentiated series is \( \sum_{n=1}^{\infty} n x^{n-1}/n^p \).

Differentiate the General Term

For each positive integer \(n\), the derivative of \(x^n/n^p\) is \(n x^{n-1}/n^p\). Since \(n/n^p=1/n^{p-1}\), the differentiated series is

\sum_{n=1}^{\infty}\frac{n x^{n-1}}{n^p} = \sum_{n=1}^{\infty}\frac{x^{n-1}}{n^{p-1}}.

The differentiation has reduced the power of \(n\) in the denominator by one. This is the source of the endpoint change. For example, at \(x=1\), the original terms are \(1/n^p\), while the differentiated terms are \(1/n^{p-1}\).

Theorem: A power series can be differentiated term by term at every point strictly inside its radius of convergence. The differentiated power series has the same radius of convergence, and its sum equals the derivative of the original sum throughout that open interval. This theorem does not, by itself, assert that the equality holds at an endpoint.

For the series here, the coefficients are \(1/n^p\). For every fixed real \(p\), their \(n\)th roots tend to \(1\), since \( (1/n^p)^{1/n}=n^{-p/n}\to 1 \). Thus the radius of convergence is \(1\). The termwise differentiation theorem therefore gives, for \(-1<x<1\),

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

The right side also has radius \(1\). To determine its behavior at \(x=1\) and \(x=-1\), substitute those values into the series and use the endpoint criteria.

Endpoint Criteria Needed for the Comparison

At \(x=1\), the original series becomes the \(p\)-series \( \sum 1/n^p \). It converges exactly when \(p>1\). At \(x=-1\), it becomes an alternating series, \( \sum (-1)^n/n^p \). It converges exactly when \(p>0\): for \(p>0\), the term magnitudes decrease to zero; for \(p\leq 0\), the terms do not tend to zero. When \(p>1\), the alternating series is absolutely convergent; when \(0<p\leq1\), it is conditionally convergent.

After differentiation, the endpoint series have exponent \(p-1\). At \(x=1\), the resulting \(p\)-series converges exactly when \(p-1>1\), or \(p>2\). At \(x=-1\), the series alternates with term magnitudes \(1/n^{p-1}\). It converges exactly when \(p-1>0\), or \(p>1\). It is absolutely convergent if \(p>2\), and conditionally convergent if \(1<p\leq2\).

Endpoint Verdicts Side by Side

Series and endpoint	Form after substitution	Convergence condition
Original at \(x=1\)	\(\sum 1/n^p\)	Converges if and only if \(p>1\)
Original at \(x=-1\)	\(\sum (-1)^n/n^p\)	Converges if and only if \(p>0\)
Differentiated at \(x=1\)	\(\sum 1/n^{p-1}\)	Converges if and only if \(p>2\)
Differentiated at \(x=-1\)	\(\sum (-1)^{n-1}/n^{p-1}\)	Converges if and only if \(p>1\)

The sign at the left endpoint changes from \((-1)^n\) to \((-1)^{n-1}\), but multiplying every term by \(-1\) does not affect convergence. The important change is the exponent \(p-1\), which makes the differentiated series require a larger value of \(p\) at both endpoints.

Why the Exponent Shift Gives These Results

The calculation of the differentiated term is direct, but the endpoint conclusions follow from familiar convergence tests. At \(x=1\), differentiating changes a positive-term \(p\)-series into one with exponent \(p-1\). Since a \(p\)-series converges only when its exponent is greater than \(1\), the condition becomes \(p-1>1\), hence \(p>2\).

At \(x=-1\), the differentiated terms alternate. The alternating-series criterion requires the magnitudes to decrease to zero. The magnitudes are \(1/n^{p-1}\), which tend to zero and decrease when \(p-1>0\). If \(p-1\leq0\), they do not tend to zero, so the series diverges by the term test. This gives the condition \(p>1\). Absolute convergence at the left endpoint requires the corresponding positive series to converge, which means \(p-1>1\), or \(p>2\).

Differentiate a term.
\(\frac{d}{dx}(x^n/n^p)=x^{n-1}/n^{p-1}\).

Substitute \(x=1\).
The signs are all positive, so convergence is determined by the \(p\)-series exponent \(p-1\).

Substitute \(x=-1\).
The signs alternate, and the term magnitudes are \(1/n^{p-1}\).

Apply the relevant test.
The positive endpoint needs exponent greater than \(1\); the alternating endpoint needs exponent greater than \(0\).

Worked Examples

Worked Example: \(p=2\)

Consider the original series \( \sum x^n/n^2 \). Its termwise derivative is obtained one term at a time:

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

At \(x=1\), the original series is \( \sum 1/n^2 \), which converges absolutely because \(2>1\). The differentiated series is \( \sum 1/n \), the harmonic series, and diverges.

At \(x=-1\), the original series is \( \sum (-1)^n/n^2 \). Its absolute-value series is \( \sum 1/n^2 \), so it converges absolutely. The differentiated series becomes \( \sum (-1)^{n-1}/n \). Its terms alternate, their magnitudes \(1/n\) decrease to zero, and its absolute-value series is the divergent harmonic series. Thus it converges conditionally.

Check: \(p=2\) satisfies \(p>1\), so both original endpoint series converge absolutely. For the derivative, \(p\) is not greater than \(2\), but it is greater than \(1\); the positive endpoint diverges and the negative endpoint converges conditionally, as found directly.

Worked Example: \(p=\frac{3}{2}\)

For this value, differentiating gives

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

At \(x=1\), the original series is \( \sum 1/n^{3/2} \), which converges absolutely because \(3/2>1\). The differentiated series is \( \sum 1/n^{1/2} \), which diverges because \(1/2\leq1\).

At \(x=-1\), the original series is \( \sum (-1)^n/n^{3/2} \), absolutely convergent because \( \sum 1/n^{3/2} \) converges. The differentiated series is \( \sum (-1)^{n-1}/n^{1/2} \). Its magnitudes decrease to zero, so the alternating series converges. Its absolute-value series \( \sum 1/n^{1/2} \) diverges; therefore the convergence is conditional.

Check: \(1<3/2<2\). The endpoint criteria predict that the derivative diverges at \(x=1\) and converges conditionally at \(x=-1\), matching the substitutions.

Worked Example: \(p=3\)

Differentiating the general term gives

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

At \(x=1\), the original series is \( \sum 1/n^3 \), and the differentiated series is \( \sum 1/n^2 \). Both converge absolutely because their exponents, \(3\) and \(2\), are greater than \(1\).

At \(x=-1\), the original series is \( \sum (-1)^n/n^3 \), and the differentiated series is \( \sum (-1)^{n-1}/n^2 \). In each case the series of absolute values converges: the corresponding exponents are \(3\) and \(2\). Both endpoint series therefore converge absolutely.

Check: \(p=3>2\), so all four endpoint series—original and differentiated, at both endpoints—should converge absolutely. The direct substitutions give exactly that result.

A Common Mistake: Applying the Derivative Theorem at an Endpoint

A tempting but incorrect argument is: “The derivative series is obtained by differentiating the original series, so its sum at \(x=1\) must be the derivative of the original sum there.” The term-by-term differentiation theorem only guarantees equality for points strictly inside the radius of convergence. At an endpoint, the differentiated series may diverge even when the original series converges there.

For a concrete case, take \(p=2\). The original series at \(x=1\) is \( \sum 1/n^2 \), which converges. The differentiated series at \(x=1\) is \( \sum 1/n \), which diverges. Thus convergence of the original endpoint series does not establish convergence of the differentiated endpoint series, much less the derivative identity there.

Takeaway: Use term-by-term differentiation inside the radius of convergence. At each endpoint, substitute into the differentiated series and test that numerical series independently.

Why This Comparison Matters

Differentiation can make boundary convergence more restrictive by reducing the coefficient exponent from \(p\) to \(p-1\). For the positive endpoint, the threshold shifts from \(p>1\) to \(p>2\). For the negative endpoint, it shifts from \(p>0\) to \(p>1\). Alternation helps at \(x=-1\), but it cannot compensate when the terms fail to approach zero.

This comparison also shows why endpoint conclusions should be recorded separately. For \(1<p\leq2\), for example, the differentiated series converges conditionally at \(x=-1\) but diverges at \(x=1\). Knowing only that the series converges at one boundary point would miss that distinction.

Check Your Understanding

For each question, distinguish the original series from its termwise differentiated series and test the two endpoints independently.

What is the termwise derivative of \(x^n/n^p\), and what exponent appears in its denominator?
For which values of \(p\) does the original series converge at \(x=1\)?
For which values of \(p\) does the differentiated series converge conditionally at \(x=-1\)?
When \(p=1\), determine the convergence or divergence of both original endpoint series and both differentiated endpoint series.
Why does the term-by-term differentiation theorem not by itself settle convergence or derivative equality at \(x=1\) or \(x=-1\)?