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Endpoint Analysis · Tutorial 745 of 998

Parity And Symmetry In Endpoint Behavior

Use coefficient parity to compare endpoint series, identify matching convergence behavior, and recognize when even–odd cancellation changes an endpoint verdict.

Intermediate 9 min read

What You'll Learn

  • Write the two endpoint series of a power series centered at zero.
  • Determine how vanishing odd-index or even-index coefficients affect endpoint terms.
  • Prove when endpoint convergence and sums must match or have opposite signs.
  • Separate endpoint partial sums into their even and odd contributions.
  • Recognize how mixed parity terms can produce different endpoint verdicts.

Why parity matters at the endpoints

A power series centered at zero has the form \(\sum_{n=0}^{\infty}a_nx^n\). At two opposite points, \(x=R\) and \(x=-R\), the even powers keep the same sign while the odd powers change sign. This simple sign change can determine whether the endpoint series agree, are negatives of each other, or behave differently because their terms combine in different ways.

Here we consider a power series with finite positive radius of convergence \(R\). The question is not whether it converges inside that radius, but how its coefficients relate the two boundary series. In particular, coefficients that vanish on one parity of indices create a strong symmetry between the endpoints.

Key idea: Replacing \(R\) by \(-R\) leaves every even-index term unchanged and reverses the sign of every odd-index term.

The two endpoint series

For \(\sum_{n=0}^{\infty}a_nx^n\), substituting the two endpoints gives the series

$$ \sum_{n=0}^{\infty}a_nR^n \qquad\text{and}\qquad \sum_{n=0}^{\infty}a_n(-R)^n. $$

Since \((-R)^n=(-1)^nR^n\), the first endpoint has terms \(a_nR^n\), while the second has terms \((-1)^na_nR^n\). Thus, the second endpoint keeps the even-index terms and changes the signs of the odd-index terms.

Definition: The even-index contribution at the endpoints is the series formed from terms \(a_{2k}R^{2k}\). The odd-index contribution is the series formed from terms \(a_{2k+1}R^{2k+1}\). The corresponding partial sums through index \(N\) are $$ E_N=\sum_{\substack{0\leq 2k\leq N}}a_{2k}R^{2k}, \qquad O_N=\sum_{\substack{0\leq 2k+1\leq N}}a_{2k+1}R^{2k+1}. $$

These definitions keep track of the original index \(n\), even when \(N\) is odd or even. If \(S_N^+\) and \(S_N^-\) denote the endpoint partial sums through index \(N\), then

$$ S_N^+=E_N+O_N, \qquad S_N^-=E_N-O_N. $$

The equations are finite identities: no convergence assumption is needed to use them for a fixed \(N\). Whether they imply a relationship between the infinite sums depends on which contributions converge.

When one parity of coefficients vanishes

If every odd-index coefficient is zero, only even powers remain. Substitution of either endpoint gives the same terms in the same order. Therefore the two endpoint partial sums are equal for every \(N\), and the endpoint series either both converge or both diverge. If they converge, their sums are equal.

If every even-index coefficient is zero, only odd powers remain. Substitution of \(-R\) changes the sign of every term. The partial sums at the endpoints are negatives of one another for every \(N\). Thus the endpoint series again either both converge or both diverge; when they converge, their sums are negatives.

Theorem: For a power series \(\sum_{n=0}^{\infty}a_nx^n\) with finite positive radius \(R\):
  • If \(a_{2k+1}=0\) for every \(k\), the endpoint series at \(R\) and \(-R\) have identical partial sums and identical convergence behavior.
  • If \(a_{2k}=0\) for every \(k\), their endpoint partial sums are negatives, so they have identical convergence behavior and opposite sums when they converge.

Why the parity rule works

The proof follows from the finite partial sums, which avoids making any assumptions about infinite sums before they are justified.

1
Substitute the endpoints.
At \(R\), the \(n\)th term is \(a_nR^n\). At \(-R\), it is \((-1)^na_nR^n\).
2
Separate the indices.
For even \(n\), \((-1)^n=1\); for odd \(n\), \((-1)^n=-1\). Thus \(S_N^+=E_N+O_N\) and \(S_N^-=E_N-O_N\).
3
Apply the vanishing condition.
If all odd coefficients vanish, then \(O_N=0\), so \(S_N^+=S_N^-\). If all even coefficients vanish, then \(E_N=0\), so \(S_N^+=-S_N^-\).
4
Pass to convergence only after the identity is established.
Equal partial sums have the same limit behavior. Negatives of convergent partial sums converge to the negative limit; if one sequence of partial sums fails to converge, its termwise negative also fails to converge.

This also explains why absolute convergence is unaffected by changing from \(R\) to \(-R\): the absolute values of corresponding terms are identical, since \(|(-1)^na_nR^n|=|a_nR^n|\). The parity theorem is stronger for the stated cases because it compares the signed partial sums and, when they converge, the actual sums.

Worked Example: An even-only series with matching endpoint sums

Worked Example: An even-only series with matching endpoint sums

Consider the power series

$$ \sum_{k=0}^{\infty}\frac{x^{2k}}{(k+1)(k+2)}. $$

Only even powers occur, so all odd-index coefficients are zero. Its radius is \(1\): the \(2k\)th roots of the nonzero coefficients satisfy

$$ \left(\frac{1}{(k+1)(k+2)}\right)^{1/(2k)} \longrightarrow 1, $$

so the radius determined by these coefficients is \(1\). At either endpoint \(x=1\) or \(x=-1\), each term becomes

$$ \frac{(\pm1)^{2k}}{(k+1)(k+2)} = \frac{1}{(k+1)(k+2)}. $$

For \(k=0,1,2\), these terms are \(1/2\), \(1/6\), and \(1/12\). In general,

$$ \frac{1}{(k+1)(k+2)} = \frac{1}{k+1}-\frac{1}{k+2}. $$

Thus the partial sum through \(k=m\) is

$$ \sum_{k=0}^{m}\frac{1}{(k+1)(k+2)} = \left(1-\frac12\right) +\left(\frac12-\frac13\right) +\cdots +\left(\frac{1}{m+1}-\frac{1}{m+2}\right) = 1-\frac{1}{m+2}. $$

As \(m\) tends to infinity, \(1/(m+2)\) tends to zero. Therefore both endpoint series converge to \(1\). Substituting either endpoint gives the same terms, and the computed partial-sum limit confirms the matching sums.

Worked Example: An odd-only series with opposite endpoint sums

Worked Example: An odd-only series with opposite endpoint sums

Consider

$$ \sum_{k=0}^{\infty}\frac{x^{2k+1}}{(k+1)(k+2)}. $$

Only odd powers occur, so all even-index coefficients vanish. The nonzero coefficients have \(2k\)th-root scale tending to \(1\), and the series has radius \(1\). At \(x=1\), the partial sum through \(k=m\) is the same telescoping sum as in the previous example:

$$ \sum_{k=0}^{m}\frac{1^{2k+1}}{(k+1)(k+2)} = 1-\frac{1}{m+2} \longrightarrow 1. $$

At \(x=-1\), every odd power is \(-1\), so the partial sum is

$$ \sum_{k=0}^{m}\frac{(-1)^{2k+1}}{(k+1)(k+2)} = -\sum_{k=0}^{m}\left(\frac{1}{k+1}-\frac{1}{k+2}\right) = -\left(1-\frac{1}{m+2}\right) \longrightarrow -1. $$

Both endpoints converge, and their sums are opposites, as required by the odd-power sign change. The endpoint terms themselves also show the relationship directly: each term at \(-1\) is the negative of its counterpart at \(1\).

When both parities are present

If even and odd coefficients are both present, neither of the simple matching rules necessarily applies. At each finite index, the endpoint partial sums still satisfy \(S_N^+=E_N+O_N\) and \(S_N^-=E_N-O_N\), but the two contributions can combine differently. In particular, one endpoint can converge while the other diverges.

Worked Example: Mixed parity gives different endpoint verdicts

Consider

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

The coefficient roots satisfy \((1/n)^{1/n}\to1\), so the radius is \(1\). At \(x=1\), the series is \(\sum_{n=1}^{\infty}1/n\), the harmonic series. Its partial sums are unbounded; for example, each block from \(2^{j-1}+1\) through \(2^j\) contains \(2^{j-1}\) terms, each at least \(1/2^j\), so each block contributes at least \(1/2\). Therefore the endpoint series at \(1\) diverges.

At \(x=-1\), the series is \(\sum_{n=1}^{\infty}(-1)^n/n\), with partial sums beginning \(-1\), \(-1+1/2\), and \(-1+1/2-1/3\). Its term magnitudes \(1/n\) decrease to zero, so the Alternating Series Test gives convergence. Thus the endpoint at \(-1\) converges while the endpoint at \(1\) diverges. The coefficients of both parities are present, allowing the endpoint terms to combine differently.

As a check, the endpoint terms at \(1\) do not tend to zero, so divergence is necessary there. At \(-1\), the terms do tend to zero, and the decreasing alternating magnitudes provide the convergence test. These checks support the distinct endpoint verdicts.

A useful consequence when both endpoints converge

There is a further consequence of the partial-sum identities. If both endpoint series converge, then their even and odd contributions must each converge separately. Indeed, adding and subtracting the endpoint partial sums gives

$$ E_N=\frac{S_N^++S_N^-}{2}, \qquad O_N=\frac{S_N^+-S_N^-}{2}. $$

If \(S_N^+\) and \(S_N^-\) have finite limits, these expressions also have finite limits. Conversely, if both parity contributions converge, then their sum and difference converge, so both endpoint series converge. This statement concerns convergence at both endpoints together; convergence at just one endpoint does not guarantee that the even and odd contributions converge separately. The mixed-parity example illustrates that caution: one endpoint can converge through cancellation even though the separate parity contributions do not both converge.

Takeaway: Vanishing coefficients on one parity force matching convergence behavior at the two endpoints. With both parities present, use the partial-sum identities and analyze each endpoint; do not infer one endpoint's verdict from the other.

Common mistake: treating every power as if its sign changed

A frequent error is to substitute \(x=-R\) and attach a minus sign to every term. For instance, the incorrect calculation

$$ a_0+a_1R+a_2R^2+\cdots \quad\longmapsto\quad -\bigl(a_0+a_1R+a_2R^2+\cdots\bigr) $$

would reverse the constant and even-power terms as well as the odd-power terms. But the actual substitution is

$$ \sum_{n=0}^{\infty}a_n(-R)^n = a_0-a_1R+a_2R^2-a_3R^3+\cdots. $$

Only odd-index terms change sign. The correction is to write \((-R)^n=(-1)^nR^n\) before drawing conclusions. That separates the even and odd indices and makes clear whether the coefficient-vanishing condition needed for a simple endpoint comparison is actually satisfied.

Check Your Understanding

Use the endpoint partial sums and parity of the coefficients to answer these questions.

  1. If \(a_{2k+1}=0\) for every \(k\), what is the relationship between the endpoint partial sums at \(R\) and \(-R\)?
  2. If \(a_{2k}=0\) for every \(k\), what is the relationship between the endpoint sums when they converge?
  3. For a series with both even and odd coefficients, write \(S_N^+\) and \(S_N^-\) in terms of \(E_N\) and \(O_N\).
  4. Why does convergence at both endpoints imply convergence of the even and odd contributions separately?
  5. Can convergence at one endpoint alone guarantee that both parity contributions converge? Explain using the partial-sum identities.

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