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Alternation Created by the Left Endpoint

Track endpoint signs carefully and determine when left-endpoint terms converge by alternating their coefficient magnitudes.

Intermediate 9 min read

What You'll Learn

  • Substitute the left endpoint into a power series centered at an arbitrary point
  • Identify when the endpoint terms have signs determined by the parity of the index
  • State and justify the alternating-series test
  • Test left-endpoint coefficients of the form one over n to a power
  • Test left-endpoint coefficients involving one over n times log n
  • Recognize when coefficient signs cancel the alternation from the endpoint

Why the Left Endpoint Can Change the Sign Pattern

A power series may have different convergence behavior at its two boundary points. At the left endpoint, the displacement from the center is negative. Raising that displacement to successive integer powers introduces a factor of \((-1)^n\). When the coefficients at the boundary otherwise have one sign, this creates the sign pattern needed for the alternating-series test.

The sign change alone does not decide convergence. The magnitudes must also approach zero and decrease in the required way. Moreover, if the coefficients already change sign, their signs combine with the factor from the endpoint. The calculation must therefore begin with direct substitution.

Substituting the Left Endpoint

Consider a power series centered at \(a\), with positive finite radius of convergence \(R\). Its general term is \(c_n(x-a)^n\). At the left endpoint \(x=a-R\), the displacement is \(x-a=-R\), so

$$ c_n(x-a)^n = c_n(-R)^n = (-1)^n c_nR^n. $$

This identity isolates two sources of sign: \((-1)^n\), which comes from evaluating at the left endpoint, and \(c_nR^n\), which comes from the coefficient. If \(c_nR^n\geq 0\), the endpoint terms are represented as \((-1)^n b_n\), where \(b_n=c_nR^n\geq 0\).

Definition: A series with terms \((-1)^n b_n\), where \(b_n\geq0\), has signs determined by the parity of \(n\), with zero terms allowed. These nonzero terms need not alternate after zero terms are deleted. The alternating-series test applies when the magnitudes \(b_n\) also satisfy its hypotheses.

The constant term, if the series is indexed from \(n=0\), has no effect on convergence. It is just one finite term. Thus it is often convenient to state the boundary test for the terms with \(n\geq1\).

The Alternating-Series Test

Theorem: Suppose \(b_n\geq0\), the sequence \(b_n\) is nonincreasing from some index onward, and \(\lim_{n\to\infty}b_n=0\). Then the series \(\sum_{n=1}^{\infty}(-1)^n b_n\) converges. The same conclusion holds if the signs start with \(+b_1\), since multiplying every term by \(-1\) does not affect convergence.

“Nonincreasing” means \(b_{n+1}\leq b_n\); strict decrease is not required. The conditions need only hold after some fixed index, because adding or removing finitely many terms does not affect whether a series converges.

Why the Test Works

Here is a proof when the conditions hold for every \(n\). Let \(S_m=\sum_{n=1}^{m}(-1)^n b_n\) be the \(m\)th partial sum. The even partial sums satisfy

$$ S_{2m+2}-S_{2m} = -b_{2m+1}+b_{2m+2} \leq 0, $$

because \(b_{2m+2}\leq b_{2m+1}\). Thus the even partial sums are nonincreasing. They are bounded below, since

$$ S_{2m} = -b_1+(b_2-b_3)+(b_4-b_5)+\cdots +(b_{2m-2}-b_{2m-1})+b_{2m} \geq -b_1. $$

Every parenthesized difference is nonnegative, as is \(b_{2m}\). A nonincreasing sequence bounded below has a finite limit, so the even partial sums converge to some number \(L\). The odd partial sums differ from the preceding even partial sums by one term:

$$ S_{2m+1}=S_{2m}-b_{2m+1}. $$

Since \(b_{2m+1}\to0\), the odd partial sums also approach \(L\). Therefore the full sequence of partial sums converges, which proves convergence of the series. If the conditions hold only from some index onward, apply this argument to the tail; the omitted initial terms are a finite sum and cannot change convergence.

Worked Example: Coefficients \(1/n^p\)

Worked Example: A Left Endpoint with Power-Law Coefficients

For a fixed \(p>0\), consider the power series

$$ \sum_{n=1}^{\infty}\frac{(x-2)^n}{n^p\,3^n}. $$

Its coefficients are \(c_n=1/(n^p3^n)\), and its boundary displacement at the left endpoint is \(-3\). Substituting \(x=2-3=-1\) gives the \(n\)th term

$$ \frac{(-1-2)^n}{n^p\,3^n} = \frac{(-3)^n}{n^p\,3^n} = \frac{(-1)^n3^n}{n^p\,3^n} = \frac{(-1)^n}{n^p}. $$

The magnitudes are \(b_n=1/n^p\). They are positive and decreasing, because \(n+1>n\) and \(p>0\) imply \((n+1)^p>n^p\), hence \(1/(n+1)^p<1/n^p\). Also, \(1/n^p\to0\) for \(p>0\). The alternating-series test therefore shows that the series converges at \(x=-1\).

For example, when \(p=1/2\), the endpoint series begins

$$ -\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}} +\frac{1}{\sqrt{4}}-\cdots = -1+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac12-\cdots. $$

The magnitudes decrease to zero, so this particular endpoint series converges. This conclusion is about the left endpoint; it does not follow merely from the radius of convergence, nor does it automatically determine behavior at the other endpoint.

Check: Substitution back into the original \(n\)th term gives \((-3)^n/(n^{1/2}3^n)=(-1)^n/\sqrt n\), exactly the series just tested.

Worked Example: A Logarithmic Factor

Worked Example: Coefficients \(1/(n\log n)\)

The expression \(1/(n\log n)\) is defined for integers \(n\geq2\), since \(\log 1=0\). Consider

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

The center is \(1\) and the left boundary displacement is \(-2\), so the left endpoint is \(x=-1\). Its \(n\)th term becomes

$$ \frac{(-1-1)^n}{n\log n\,2^n} = \frac{(-2)^n}{n\log n\,2^n} = \frac{(-1)^n}{n\log n}. $$

Set \(b_n=1/(n\log n)\) for \(n\geq2\). These magnitudes are positive. To check that they decrease, consider \(f(t)=1/(t\log t)\) for \(t>1\). Its derivative is

$$ f'(t) = -\frac{\log t+1}{(t\log t)^2}<0 \qquad (t>1), $$

so \(f\) is decreasing and therefore \(b_{n+1}<b_n\) for integers \(n\geq2\). Also, \(n\log n\to\infty\), so \(b_n\to0\). The alternating-series test proves convergence at \(x=-1\).

The first terms make the parity pattern visible:

$$ -\frac{1}{2\log2} +\frac{1}{3\log3} -\frac{1}{4\log4} +\frac{1}{5\log5} -\cdots. $$

Check: At \(x=-1\), the original factor \((x-1)^n\) is \((-2)^n\); dividing by \(2^n\) leaves \((-1)^n\). Thus the displayed endpoint terms have the correct signs and magnitudes.

Coefficient Signs Can Cancel the Endpoint Alternation

The left endpoint supplies a factor \((-1)^n\), but that does not guarantee that the resulting terms alternate. If \(c_n\) itself has a factor \((-1)^n\), the two factors multiply to \(1\). A sign analysis must include both.

Worked Example: A Sign Change in the Coefficients

Consider the series centered at \(1\)

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

The left endpoint is \(x=1-2=-1\). Substitution gives

$$ \frac{(-1)^n(-1-1)^n}{n^2\,2^n} = \frac{(-1)^n(-2)^n}{n^2\,2^n} = \frac{(-1)^n(-1)^n2^n}{n^2\,2^n} = \frac{1}{n^2}. $$

The two parity factors cancel, so the endpoint terms are all positive, not alternating. The endpoint series is \(\sum_{n=1}^{\infty}1/n^2\), which converges by the \(p\)-series criterion with \(p=2>1\).

Check: Multiplying the coefficient \((-1)^n/(n^2 2^n)\) by the substituted power \((-2)^n\) gives \(1/n^2\) for every \(n\), confirming the cancellation.

A Common Sign Error

A frequent mistake is to substitute the left endpoint but omit the negative sign in the displacement. In the first example, an incorrect calculation might read

$$ \frac{(x-2)^n}{n^p3^n}\bigg|_{x=-1} \ \text{(incorrectly)}\ = \frac{3^n}{n^p3^n} = \frac{1}{n^p}. $$

This treats \(x-2\) as \(3\), although at \(x=-1\) it is \(-1-2=-3\). The corrected calculation is

$$ \frac{(-3)^n}{n^p3^n} = \frac{(-1)^n}{n^p}. $$

Dropping \((-1)^n\) changes the series being tested. In particular, it replaces an alternating boundary series by a positive-term series, for which the alternating-series test cannot be used.

How to Test a Left Boundary

1
Substitute the boundary.
At the left endpoint \(x=a-R\), write \((x-a)^n=(-R)^n=(-1)^nR^n\).
2
Combine all sign factors.
Include the signs of the coefficients as well as \((-1)^n\); do not assume alternation if the coefficients already change sign.
3
Identify the magnitudes.
When the terms have the form \((-1)^n b_n\), record \(b_n\geq0\) explicitly, allowing zero terms.
4
Check both test conditions.
Verify that \(b_n\) is eventually nonincreasing and tends to zero. Only then does the alternating-series test establish convergence.

The endpoint substitution explains where the alternation comes from; the test conditions explain why that alternation is enough to give convergence. If the magnitudes fail to tend to zero, the terms themselves do not tend to zero, so the series cannot converge. If the magnitudes do tend to zero but are not eventually nonincreasing, this test alone gives no conclusion.

Check Your Understanding

Answer each question by writing the endpoint terms before deciding which convergence test applies.

  1. At the left endpoint \(x=a-R\), what is the identity for \((x-a)^n\)?
  2. State the two conditions on \(b_n\) in the alternating-series test.
  3. For \(b_n=1/n^p\), why do the alternating-series test conditions hold when \(p>0\)?
  4. Why must the series involving \(1/(n\log n)\) start at \(n=2\)?
  5. If \(c_n\) includes a factor \((-1)^n\), what sign calculation should be made before labeling the left-endpoint series alternating?

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