Why Compare a Boundary Series With Its Absolute-Value Series?
A power series can converge for every point strictly inside its radius of convergence, yet behave differently at the two boundary points. At a boundary, the radius alone does not tell us whether the series converges. We must substitute the endpoint and analyze the resulting numerical series.
One important question is whether the endpoint series still converges when all its terms are replaced by their absolute values. This comparison separates three possibilities: absolute convergence, conditional convergence, and divergence.
The distinction matters because absolute convergence is a stronger guarantee than ordinary convergence. A series may converge because of cancellation between positive and negative terms even when the corresponding series of magnitudes diverges.
Definitions at a Boundary
Suppose substituting a boundary point into a power series produces the numerical series \(\sum_{n=1}^{\infty}u_n\). Its absolute-value series is \(\sum_{n=1}^{\infty}|u_n|\). The two series are related, but their convergence need not agree.
If the original series does not converge, it is divergent. Divergence of the absolute-value series alone does not establish that the original series diverges: the original series might converge conditionally.
For a power series, the procedure is endpoint-specific. Substitute one endpoint to obtain its boundary series, then form and test that series’ absolute-value series. Repeat the process separately at the other endpoint.
Why Absolute Convergence Guarantees Convergence
The implication from absolute convergence to convergence follows directly from a bound on the tails. The proof also explains why the absolute-value series is a useful comparison.
Let \(S_N=\sum_{n=1}^{N}u_n\). For \(m>N\), the difference between two partial sums is the tail \(\sum_{n=N+1}^{m}u_n\).
The triangle inequality gives \(\left|S_m-S_N\right|=\left|\sum_{n=N+1}^{m}u_n\right|\leq\sum_{n=N+1}^{m}|u_n|\).
Since \(\sum |u_n|\) converges, its tails can be made arbitrarily small. Thus, for every \(\varepsilon>0\), there is an \(N\) such that \(m>N\) implies \(\left|S_m-S_N\right|<\varepsilon\).
The partial sums \(S_N\) form a Cauchy sequence. The real numbers are complete, so the partial sums converge. Therefore, \(\sum u_n\) converges.
The converse is false: convergence of \(\sum u_n\) does not force convergence of \(\sum |u_n|\). Alternating signs can produce enough cancellation for the original series to converge while its absolute-value series diverges.
A Practical Endpoint Procedure
At a boundary point, first write the substituted series explicitly. Do not decide its behavior merely by looking at the power-series coefficients or at the series’ behavior at the other endpoint.
- Substitute the endpoint into every occurrence of the variable.
- Identify the resulting term \(u_n\) and the original boundary series \(\sum u_n\).
- Form the absolute-value series \(\sum |u_n|\) and determine whether it converges.
- If the absolute-value series converges, the endpoint series converges absolutely.
- If the absolute-value series diverges, test the original series: it may converge conditionally or diverge.
For example, the alternating-series criterion applies to \(\sum (-1)^{n-1}b_n\) when \(b_n\geq 0\), the sequence \(b_n\) is nonincreasing, and \(b_n\) tends to zero. Under those conditions the alternating series converges. To decide whether that convergence is conditional, its absolute-value series must still be checked.
Worked Endpoint Classifications
Worked Example: A Conditional Endpoint and a Divergent Endpoint
Consider the power series \(\sum_{n=1}^{\infty}\frac{x^n}{n}\), whose candidate boundary points are \(x=-1\) and \(x=1\). Substitute each point separately.
At \(x=-1\), the terms become \[ \frac{(-1)^n}{n}, \] so the boundary series is \(\sum_{n=1}^{\infty}\frac{(-1)^n}{n}\). The magnitudes are \(b_n=1/n\). They are positive, decrease as \(n\) increases, and tend to zero. The alternating-series criterion therefore gives convergence of the original boundary series.
Now take absolute values: \[ \left|\frac{(-1)^n}{n}\right|=\frac{1}{n}. \] The absolute-value series is \(\sum_{n=1}^{\infty}\frac{1}{n}\), the harmonic series, which diverges. Thus the endpoint \(x=-1\) is a point of conditional convergence.
At \(x=1\), substitution gives \[ \frac{1^n}{n}=\frac{1}{n}. \] The boundary series is the harmonic series, so it diverges. Its absolute-value series is again \(\sum_{n=1}^{\infty}1/n\), which diverges as well. The endpoint \(x=1\) is divergent, not conditionally convergent.
Check the classifications by returning to the substituted terms: at \(-1\), the original terms alternate but their magnitudes form the divergent harmonic series; at \(1\), both the original and absolute-value series are harmonic. Therefore the first endpoint is conditional and the second is divergent.
Worked Example: Absolute Convergence at Both Endpoints
Consider \(\sum_{n=1}^{\infty}\frac{(x-1)^n}{n^2}\), with candidate endpoints \(x=0\) and \(x=2\). At \(x=0\), \(x-1=-1\), so the substituted term is \[ \frac{(-1)^n}{n^2}. \] Its absolute value is \(1/n^2\). The absolute-value series \(\sum_{n=1}^{\infty}1/n^2\) is a convergent \(p\)-series, with \(p=2>1\). The endpoint series at \(x=0\) therefore converges absolutely.
At \(x=2\), \(x-1=1\), so the substituted term is \[ \frac{1^n}{n^2}=\frac{1}{n^2}. \] Its absolute value remains \(1/n^2\), and the same convergent \(p\)-series establishes absolute convergence at \(x=2\).
Check the substitutions: the endpoint terms are \((-1)^n/n^2\) and \(1/n^2\), respectively. In both cases their magnitudes are exactly \(1/n^2\). Since that absolute-value series converges, the theorem on absolute convergence confirms convergence of each original endpoint series.
Worked Example: Conditional Convergence at One Endpoint, Divergence at the Other
Consider \(\sum_{n=1}^{\infty}\frac{(-1)^n(x-3)^n}{\sqrt n}\), with candidate endpoints \(x=2\) and \(x=4\). At \(x=4\), \(x-3=1\), and the term becomes \[ \frac{(-1)^n}{\sqrt n}. \] The magnitudes \(b_n=1/\sqrt n\) are positive, decrease as \(n\) increases, and tend to zero. The alternating-series criterion shows that the boundary series converges.
Taking absolute values removes the alternating sign: \[ \left|\frac{(-1)^n}{\sqrt n}\right|=\frac{1}{\sqrt n}. \] The absolute-value series is the \(p\)-series \(\sum_{n=1}^{\infty}1/n^{1/2}\), which diverges because \(p=1/2\leq 1\). Therefore the endpoint \(x=4\) is conditionally convergent.
At \(x=2\), \(x-3=-1\). Substitution gives \[ \frac{(-1)^n(-1)^n}{\sqrt n} = \frac{(-1)^{2n}}{\sqrt n} = \frac{1}{\sqrt n}, \] because \((-1)^{2n}=1\). This is the same divergent \(p\)-series with \(p=1/2\). Its terms do tend to zero, but that is not sufficient for convergence; the \(p\)-series diverges. The absolute-value series is also \(\sum 1/\sqrt n\), so this endpoint is divergent.
Check the endpoint substitutions: \(x=4\) leaves the factor \((-1)^n\), while \(x=2\) contributes a second factor \((-1)^n\) that cancels it. Thus the first boundary series alternates and converges conditionally, whereas the second has positive terms and diverges.
A Common Mistake
A frequent error is to conclude that an original series diverges as soon as its absolute-value series diverges. The first example shows why that reasoning fails. At \(x=-1\),
It would be incorrect to infer from this that \(\sum_{n=1}^{\infty}(-1)^n/n\) diverges. Its terms alternate, their magnitudes decrease to zero, and the alternating-series criterion proves that the original series converges. The correct conclusion is that it converges conditionally.
The logical direction of the absolute-convergence theorem is important: convergence of \(\sum |u_n|\) guarantees convergence of \(\sum u_n\), but divergence of \(\sum |u_n|\) leaves two possibilities. The original series may converge conditionally, or it may diverge. Test the original series to distinguish them.
Why the Distinction Matters
At the boundary of a power series, the same coefficient pattern can produce different behavior at opposite endpoints. Substituting the endpoints may preserve an alternating sign at one endpoint and cancel it at the other. Comparing with the absolute-value series reveals whether convergence depends on that cancellation.
Absolute convergence is the stronger conclusion: it guarantees ordinary convergence and does not rely on cancellation between signs. Conditional convergence is still genuine convergence, but it occurs even though the magnitudes alone do not sum to a finite value. Divergence means the original boundary series itself fails to converge.
Check Your Understanding
For each question, identify what must be checked before assigning an endpoint verdict.
- If a boundary series converges and its absolute-value series converges, how should the endpoint be classified?
- If an absolute-value series diverges, why is it not yet valid to say that the original boundary series diverges?
- At \(x=-1\), what boundary series results from \(\sum_{n=1}^{\infty}x^n/n\), and what is the status of its absolute-value series?
- For \(\sum_{n=1}^{\infty}(-1)^n(x-3)^n/\sqrt n\), what happens to the alternating signs at \(x=2\)?
- What three conditions allow the alternating-series criterion to establish convergence of \(\sum(-1)^{n-1}b_n\)?