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Endpoint Analysis · Tutorial 729 of 998

P-Series Coefficients at the Endpoints

Learn how the exponent \(p\) determines convergence at both endpoints of a power series with coefficients \(1/n^p\).

Intermediate 10 min read

What You'll Learn

  • Substitute each boundary value into a power series with coefficients \(1/n^p\).
  • Apply the p-series threshold to determine convergence at the positive boundary.
  • Test convergence at the negative boundary when substitution produces alternating signs.
  • Distinguish absolute convergence, conditional convergence, and divergence in this setting.
  • Write the resulting interval of convergence, including its endpoint brackets.

Why the Endpoints Need Their Own Tests

A power series can converge at every point strictly inside its radius of convergence and still behave differently at the two boundary points. When its coefficients involve \(1/n^p\), the endpoint tests reduce to familiar kinds of numerical series: a p-series at one end and, when the signs alternate, an alternating series at the other.

The exponent \(p\) controls the size of the terms. At the positive boundary, the terms are positive and the p-series threshold gives the verdict. At the negative boundary, the signs alternate, so convergence can occur even when the corresponding series of absolute values diverges.

A Model Power Series and Its Boundary Values

Consider the model power series

$$ \sum_{n=1}^{\infty}\frac{1}{n^p}\left(\frac{x-c}{R}\right)^n, \qquad R>0, $$

where \(c\) is the center, \(R\) is the radius, and \(p\) is real. The index starts at \(n=1\), so every coefficient \(1/n^p\) is defined. At the right boundary, \(x=c+R\), the factor \((x-c)/R\) equals \(1\). At the left boundary, \(x=c-R\), it equals \(-1\). Substitution therefore gives these two endpoint series:

$$ x=c+R:\quad \sum_{n=1}^{\infty}\frac{1}{n^p}, \qquad x=c-R:\quad \sum_{n=1}^{\infty}\frac{(-1)^n}{n^p}. $$

The first is a p-series. The second has alternating signs, but its absolute values are again the terms of a p-series. The model has radius \(R\): the \(n\)th root of the coefficient magnitude \(1/(n^pR^n)\) tends to \(1/R\), since \(n^{-p/n}\) tends to \(1\). The radius identifies the candidate boundaries; the endpoint substitutions determine whether either boundary is included.

Definition: A series converges absolutely if the series of absolute values of its terms converges. It converges conditionally if it converges but does not converge absolutely.

The p-Series Threshold

The p-series \(\sum_{n=1}^{\infty}1/n^p\) converges exactly when \(p>1\). A short grouping argument explains the threshold without needing to estimate individual terms by an integral.

1
Group terms in blocks.
For each integer \(k\geq 0\), group the terms with indices \(2^k\leq n<2^{k+1}\). There are \(2^k\) terms in this block.
2
Bound each block.
For \(p>0\), each term in the block is at most \(2^{-kp}\) and at least \(2^{-(k+1)p}\). Thus the block sum lies between \(2^{-p}2^{k(1-p)}\) and \(2^{k(1-p)}\).
3
Decide whether the blocks can sum.
If \(p>1\), the upper bounds form a convergent geometric series with ratio \(2^{1-p}<1\), so the p-series converges. If \(0<p\leq1\), the lower bound for each block is at least \(2^{-p}\), so infinitely many blocks contribute at least that much and the p-series diverges. If \(p\leq0\), then \(1/n^p=n^{-p}\geq1\), so the terms do not tend to zero and the series diverges.

This proves the threshold for every real \(p\). In particular, the boundary test at \(x=c+R\) is already settled: it converges for \(p>1\) and diverges for \(p\leq1\).

What Changes at the Negative Boundary?

At \(x=c-R\), the endpoint series is \(\sum (-1)^n/n^p\). If \(p>0\), the positive magnitudes \(1/n^p\) decrease to zero. The alternating-series criterion says that a series with alternating signs converges when its positive magnitudes decrease to zero. Here is a brief justification for this particular series.

Write \(a_n=1/n^p\), so \(a_n>0\), \(a_{n+1}<a_n\), and \(a_n\to0\). For its even partial sums, changing the overall sign does not affect convergence, and the relevant grouped sums are

$$ T_m=(a_1-a_2)+(a_3-a_4)+\cdots+(a_{2m-1}-a_{2m}). $$

Each added pair is positive, so \(T_m\) is increasing. Also \(T_m\leq a_1\), because the terms removed from \(a_1\) in the expression below are nonnegative:

$$ T_m=a_1-(a_2-a_3)-(a_4-a_5)-\cdots-(a_{2m-2}-a_{2m-1})-a_{2m}\leq a_1. $$

Thus the even partial sums converge, by monotone boundedness. An odd partial sum differs from the preceding even partial sum by one term whose magnitude tends to zero. The odd partial sums therefore approach the same limit, and the full alternating series converges.

For \(p>0\), this establishes convergence at the left boundary. Whether it is absolute depends on the p-series of absolute values, \(\sum 1/n^p\). If \(p>1\), that p-series converges, so the endpoint convergence is absolute. If \(0<p\leq1\), the p-series diverges, so the alternating endpoint converges only conditionally. For \(p\leq0\), the terms \((-1)^n/n^p\) do not tend to zero in magnitude, so the alternating series diverges by the term test.

Endpoint Classification

ExponentRight boundary \(x=c+R\)Left boundary \(x=c-R\)Interval of convergence
\(p>1\)Converges absolutelyConverges absolutely\([c-R,c+R]\)
\(0<p\leq1\)DivergesConverges conditionally\([c-R,c+R)\)
\(p\leq0\)DivergesDiverges\((c-R,c+R)\)

The interval entries assume the power series converges throughout the interior \((c-R,c+R)\), as it does for a power series of radius \(R\). The table records endpoint behavior only through the brackets: a square bracket includes the boundary point, while a parenthesis excludes it.

Worked Examples

Worked Example: A Square-Summable Boundary

Classify the endpoints of

$$ \sum_{n=1}^{\infty}\frac{1}{n^2}\left(\frac{x-1}{4}\right)^n. $$

Here \(c=1\), \(R=4\), and \(p=2\). The candidate endpoints are \(1-4=-3\) and \(1+4=5\).

At \(x=5\), \((x-1)/4=(5-1)/4=1\), so the series becomes

$$ \sum_{n=1}^{\infty}\frac{1}{n^2}. $$

Since \(p=2>1\), this p-series converges. At \(x=-3\), \((x-1)/4=(-3-1)/4=-1\), so the series becomes

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

Its absolute-value series is \(\sum 1/n^2\), which converges by the same \(p=2>1\) test. Thus both endpoints converge absolutely. Including the interior, the interval is \([-3,5]\). Substituting the endpoints back into the original factor confirms that the respective factors are \(-1\) and \(1\), as required.

Worked Example: The Harmonic Threshold

Classify the endpoints of

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

Here \(c=-2\), \(R=3\), and \(p=1\). The endpoints are \(-2-3=-5\) and \(-2+3=1\).

At \(x=1\), the factor is \((1+2)/3=1\), producing the harmonic series \(\sum 1/n\). Since \(p=1\leq1\), it diverges. At \(x=-5\), the factor is \((-5+2)/3=-1\), producing \(\sum (-1)^n/n\). Its terms alternate, and their magnitudes \(1/n\) decrease to zero, so the alternating series converges. Its absolute-value series is \(\sum 1/n\), which diverges. Therefore the left endpoint converges conditionally, while the right endpoint diverges. The interval is \([-5,1)\). Substitution checks the endpoint factors: \(-1\) at \(-5\) and \(1\) at \(1\).

Worked Example: Coefficients That Do Not Tend to Zero

Classify the endpoints of

$$ \sum_{n=1}^{\infty}\left(\frac{x-4}{2}\right)^n. $$

This is the case \(p=0\), since \(1/n^0=1\), with center \(c=4\) and radius \(R=2\). Its endpoints are \(4-2=2\) and \(4+2=6\).

At \(x=6\), the factor is \((6-4)/2=1\). The terms are all \(1\), so they do not tend to zero and the series diverges. At \(x=2\), the factor is \((2-4)/2=-1\). The series is \(\sum (-1)^n\), whose terms have magnitude \(1\); they do not tend to zero, so this series also diverges. Neither endpoint is included, and the interval is \((2,6)\). Substitution verifies the factors \(1\) and \(-1\) directly.

A Common Endpoint Error

A frequent mistake is to see alternating signs at the left boundary and conclude immediately that the series converges. Alternation alone is not enough; for example, at \(p=0\), the terms \((-1)^n\) alternate but do not approach zero.

Correction: Before applying the alternating-series criterion, check that the magnitudes decrease to zero. For \(1/n^p\), this condition holds exactly when \(p>0\). When \(p\leq0\), the term test rules out convergence at either endpoint.

Another error is to use the right-boundary p-series verdict at both endpoints. The absolute-value series is the same at both ends, but the actual left-boundary series can converge conditionally when \(0<p\leq1\). Substitute each endpoint separately before deciding whether to include it.

Why the Classification Matters

The exponent divides endpoint behavior into three distinct regimes. For \(p>1\), the coefficients decrease quickly enough for the p-series itself to converge, so both endpoints are included. For \(0<p\leq1\), the positive-boundary p-series diverges, but the negative boundary gains convergence from alternating signs. For \(p\leq0\), the terms fail to approach zero at either endpoint, so neither boundary can converge.

This distinction is useful whenever a power series has reciprocal-power coefficients: it turns endpoint analysis into a direct check of the exponent and the signs after substitution. It also shows why knowing the radius alone cannot settle endpoint inclusion.

Check Your Understanding

For each question, use the endpoint substitutions and the convergence conditions developed above.

  1. For \(\sum_{n=1}^{\infty}((x-3)/2)^n/n^{3/2}\), which endpoints converge, and what is the interval of convergence?
  2. For \(\sum_{n=1}^{\infty}((x+1)/5)^n/n^{1/2}\), classify each endpoint as divergent, absolutely convergent, or conditionally convergent.
  3. Why does \(\sum (-1)^n/n^p\) fail the term test when \(p\leq0\)?
  4. For \(0<p\leq1\), why is the left-boundary convergence conditional rather than absolute?
  5. At what values of \(p\) does the model series converge at both endpoints?

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