Endpoint Behavior After Termwise Differentiation and Integration

Why Endpoints Need Their Own Tests

Inside a power series’ radius of convergence, differentiation and integration can be carried out term by term. At an endpoint, however, the resulting series must be tested directly: convergence of the original series there does not guarantee convergence of its differentiated series, and divergence of the original series does not prevent its integrated series from converging.

The series for \(\ln(1+x)\) and \(\arctan x\) give clear examples. Both have radius of convergence \(1\), but their endpoint behavior is not identical. Differentiating either series removes a denominator from its coefficients; integrating adds another. Those changes can determine whether the endpoint terms tend to zero and whether the series converges.

Key principle: A power series converges absolutely at every point strictly inside its radius and diverges outside it. Its behavior at the boundary is not determined by the radius alone. At each endpoint, inspect the resulting numerical series.

The Two Starting Series

For \(|x|<1\), the power series we will compare are

\ln(1+x)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^n}{n}, \qquad \arctan x=\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{2n+1}.

Each identity holds throughout the open interval \((-1,1)\). The endpoint values must be considered separately. For \(\ln(1+x)\), the series converges at \(x=1\) but diverges at \(x=-1\). For \(\arctan x\), it converges at both endpoints. In these cases, endpoint convergence is conditional rather than absolute.

A series is absolutely convergent if the series of absolute values converges. It is conditionally convergent if it converges but its series of absolute values diverges. A necessary condition for \(\sum a_n\) to converge is that \(a_n\to0\). If its terms do not tend to zero, the series diverges.

Series	At the left endpoint	At the right endpoint
\(\ln(1+x)\)	\(x=-1\): diverges like \(-\sum 1/n\)	\(x=1\): converges conditionally to \(\ln 2\)
\(\arctan x\)	\(x=-1\): converges conditionally to \(-\pi/4\)	\(x=1\): converges conditionally to \(\pi/4\)

What Differentiation Does at the Endpoints

Within \((-1,1)\), termwise differentiation gives

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

In each case, the factor in the denominator of the original coefficient has disappeared. At the endpoints, this leaves terms that do not approach zero. The differentiated series therefore diverge at both endpoints, even where the original series converged.

Worked Example: Differentiating the Series for \(\ln(1+x)\)

Start with the \(n\)th term, \(\frac{(-1)^{n+1}x^n}{n}\). Differentiating it gives

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

Thus, for \(|x|<1\), the differentiated series is \(\sum_{n=1}^{\infty}(-1)^{n+1}x^{n-1}\). At \(x=1\), its terms are \(1,-1,1,-1,\ldots\), which do not tend to zero. At \(x=-1\), its \(n\)th term is

(-1)^{n+1}(-1)^{n-1}=(-1)^{2n}=1.

These terms also do not tend to zero. Therefore the differentiated series diverges at both endpoints. This agrees with its interior sum: the geometric series equals \(1/(1+x)\) for \(|x|<1\), which is undefined at \(x=-1\) and does not supply a convergent endpoint series at \(x=1\). Check: multiplying the geometric series by \(1+x\) gives \(1\) in the open interval, confirming its stated sum there.

Worked Example: Differentiating the Series for \(\arctan x\)

Differentiate the general term of the arctangent series:

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

The resulting series is \(\sum_{n=0}^{\infty}(-1)^n x^{2n}\) for \(|x|<1\). At \(x=1\), it becomes \(1-1+1-1+\cdots\); at \(x=-1\), it is the same because \((-1)^{2n}=1\). In both cases the terms have absolute value \(1\), so the terms fail to approach zero and the series diverges.

For \(|x|<1\), this is a geometric series with ratio \(-x^2\), so its sum is \(1/(1+x^2)\). Differentiating \(\arctan x\) gives the same function. Check: \((1+x^2)\sum_{n=0}^{\infty}(-1)^n x^{2n}=1\) inside the interval, as required.

What Integration Does at the Endpoints

Termwise integration, using zero as the lower limit, gives two different series:

\int_0^x \ln(1+t)\,dt = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^{n+1}}{n(n+1)}, \qquad \int_0^x \arctan t\,dt = \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+2}}{(2n+1)(2n+2)}.

These identities initially follow for \(|x|<1\), where termwise integration is valid. Integration adds a power of \(x\) and divides each coefficient by the new exponent. At either endpoint, that produces denominators of order \(n^2\), so the integrated series converge absolutely there.

Worked Example: Integrating the Series for \(\ln(1+x)\)

Integrate each term from \(0\) to \(x\):

\int_0^x\frac{(-1)^{n+1}t^n}{n}\,dt = \frac{(-1)^{n+1}}{n}\left[\frac{t^{n+1}}{n+1}\right]_0^x = \frac{(-1)^{n+1}x^{n+1}}{n(n+1)}.

At \(x=-1\), the \(n\)th term is \(1/[n(n+1)]\); at \(x=1\), its absolute value is also \(1/[n(n+1)]\). The endpoint series converge absolutely because

\frac{1}{n(n+1)}=\frac1n-\frac{1}{n+1}, \qquad \sum_{n=1}^{N}\frac{1}{n(n+1)} = 1-\frac{1}{N+1} \longrightarrow 1.

The integrated function can also be evaluated directly. An antiderivative of \(\ln(1+x)\) is \((1+x)\ln(1+x)-(1+x)\), so

\int_0^x\ln(1+t)\,dt = (1+x)\ln(1+x)-x.

At \(x=1\), this is \(2\ln 2-1\). At \(x=-1\), use \(\lim_{u\to0^+}u\ln u=0\), giving \(1\). The series at \(x=-1\) gives the same value by telescoping. Check: differentiating \((1+x)\ln(1+x)-x\) gives \(\ln(1+x)+1-1=\ln(1+x)\), and its value at \(x=0\) is \(0\).

Worked Example: Integrating the Series for \(\arctan x\)

Integrating a general term gives

\int_0^x\frac{(-1)^n t^{2n+1}}{2n+1}\,dt = \frac{(-1)^n x^{2n+2}}{(2n+1)(2n+2)}.

At either \(x=1\) or \(x=-1\), the absolute value of the \(n\)th term is \(1/[(2n+1)(2n+2)]\). For \(n\geq1\), this is at most \(1/(4n^2)\), and \(\sum_{n=1}^{\infty}1/n^2\) converges. Thus both endpoint series converge absolutely.

To evaluate them, integration by parts gives, for \(x\) in the interval,

\int_0^x\arctan t\,dt = x\arctan x-\frac12\ln(1+x^2).

At \(x=1\), the value is \(\pi/4-\frac12\ln 2\). At \(x=-1\), it is also \(\pi/4-\frac12\ln 2\), since \(x\arctan x=\pi/4\) there. Check: differentiating the expression gives \(\arctan x+x/(1+x^2)-x/(1+x^2)=\arctan x\), and its value at zero is \(0\).

Why the Endpoint Results Change

The comparison is explained by the coefficients. Differentiation multiplies the coefficient of \(x^k\) by \(k\); integration divides it by the new exponent. At the boundary, where \(|x|=1\), powers of \(x\) no longer shrink in magnitude. The coefficients then largely determine whether the terms tend to zero and whether the series converges.

For the original logarithmic and arctangent series, the denominators grow like \(n\), allowing alternating endpoint convergence in the cases shown above. Differentiation cancels those denominators, leaving terms of constant magnitude at the endpoints. Integration adds an additional factor of order \(n\) to the denominator, yielding terms comparable to \(1/n^2\), whose absolute values have a convergent sum.

Important condition: Termwise differentiation and integration of a power series are justified within its open interval of convergence. This fact does not automatically assert convergence of the resulting series at either endpoint. Endpoint claims require separate tests.

A Common Endpoint Mistake

A common error is to conclude that a differentiated series converges at an endpoint because the original series converges there. For example, the series for \(\arctan 1\) converges to \(\pi/4\), but differentiating its terms and then substituting \(x=1\) gives

1-1+1-1+\cdots,

which diverges because its terms do not approach zero. The valid termwise differentiation identity holds for \(|x|<1\), not automatically at \(x=1\). The correction is to substitute the endpoint into the new series and test that numerical series itself. The same caution applies to integration: convergence at the endpoints must be established, not assumed merely because the original series has a particular endpoint behavior.

Endpoint Comparison

Series or operation	Endpoint outcome	Reason
\(\ln(1+x)\)	Diverges at \(-1\); converges conditionally at \(1\)	Harmonic divergence at \(-1\); alternating harmonic convergence at \(1\)
Derivative of \(\ln(1+x)\)	Diverges at both endpoints	Terms do not tend to zero
\(\arctan x\)	Converges conditionally at both endpoints	Alternating reciprocal odd terms
Derivative of \(\arctan x\)	Diverges at both endpoints	Terms alternate with constant magnitude
Integral of either series from \(0\)	Converges absolutely at both endpoints	Coefficients have denominators of order \(n^2\)

Check Your Understanding

For each question, distinguish what is known inside the radius from what must be tested at the endpoints.

At which endpoint does the series for \(\ln(1+x)\) diverge, and what familiar series shows the divergence?
Why does the differentiated series for \(\arctan x\) diverge at \(x=1\)?
Write the \(n\)th term of the integrated series for \(\ln(1+x)\), with lower limit \(0\).
Why do the integrated endpoint series converge absolutely, even when an original series diverges at an endpoint?
What is the key reason that termwise differentiation inside an interval does not settle convergence at its endpoints?