Logarithmic Factors in Alternating Series

Why logarithmic factors need careful attention

A power of $n$ often determines whether a series converges, but a logarithmic factor can decide what happens at a boundary. For example, when the power of $n$ is exactly one, changing the exponent on $\log n$ changes whether the corresponding positive series converges.

For an alternating series, there are two separate questions. Does the signed series converge? Does the series of absolute values converge? The answers need not be the same: a series that converges but whose absolute-value series diverges is called conditionally convergent.

Throughout, $\log x$ means the natural logarithm, and the series begins at $n=2$, so that $\log n$ is positive. Changing or adding finitely many initial terms does not affect convergence.

Definition: For real parameters $p$ and $q$, consider $$ \sum_{n=2}^{\infty}\frac{(-1)^{n-1}}{n^p(\log n)^q}. $$ Its term magnitudes are $b_n=1/[n^p(\log n)^q]$. The series is absolutely convergent if $\sum_{n=2}^{\infty} b_n$ converges, conditionally convergent if the alternating series converges but that magnitude series diverges, and divergent if the alternating series does not converge.

First test the alternating series itself

The Alternating-Series Test says that $\sum (-1)^{n-1}b_n$ converges if $b_n\ge 0$, $b_n$ is eventually nonincreasing, and $\lim_{n\to\infty}b_n=0$. “Eventually” allows a finite number of terms before the decrease begins.

For this family, consider the corresponding function

f(x)=\frac{1}{x^p(\log x)^q},\qquad x>1.

It is positive, and differentiation gives

f'(x) =-\frac{f(x)}{x}\left(p+\frac{q}{\log x}\right).

If $p>0$, then $p+q/\log x>0$ for all sufficiently large $x$, whatever real value $q$ has. Thus $f$, and hence $b_n$, is eventually decreasing. Also $b_n\to0$: a positive power of $n$ eventually dominates any fixed power of $\log n$. The Alternating-Series Test therefore gives convergence for every $q$ when $p>0$.

If $p=0$, then $b_n=(\log n)^{-q}$. For $q>0$, these terms decrease to zero, so the alternating series converges. For $q=0$, they equal $1$; for $q<0$, they grow without bound. In both latter cases the terms do not approach zero, so the series diverges. Finally, if $p<0$, the factor $n^{-p}$ grows as a positive power of $n$, and no fixed logarithmic power can make $b_n$ tend to zero. The series again diverges by the term test.

Takeaway: The alternating series converges exactly when $p>0$, or when $p=0$ and $q>0$. In the remaining cases its terms do not tend to zero.

Classifying the absolute-value series

Absolute convergence depends on the positive series

\sum_{n=2}^{\infty}\frac{1}{n^p(\log n)^q}.

To classify it, we use the Integral Test: if $f$ is positive, continuous, and decreasing for all sufficiently large $x$, then $\sum f(n)$ and $\int f(x)\,dx$ over the corresponding infinite interval either both converge or both diverge. The eventual decrease needed here follows from the derivative calculation above whenever $p>0$, and also when $p=0,\ q>0$.

The reason the test works is the comparison between a decreasing curve and rectangles. For integers $N\le M$, a positive decreasing $f$ satisfies

\int_N^{M+1} f(x)\,dx \le \sum_{n=N}^{M}f(n) \le f(N)+\int_N^M f(x)\,dx.

On each interval $[n,n+1]$, the curve lies at or below $f(n)$, giving the left inequality after summing. For the right inequality, when $n\ge N+1$, $f(n)\le\int_{n-1}^{n}f(x)\,dx$; adding the first term $f(N)$ gives the result. These inequalities show that the partial sums are bounded precisely when the improper integral is finite.

For the integral, set $t=\log x$. Since $x=e^t$ and $dx=e^t\,dt$, we obtain

\int \frac{dx}{x^p(\log x)^q} =\int e^{(1-p)t}t^{-q}\,dt.

This expression gives three regimes. If $p>1$, the exponential factor decays, and the integral converges for every real $q$. If $p<1$, the exponential factor grows and the integral diverges whenever the Integral Test applies. To see why a logarithmic power cannot reverse either conclusion, note that for any $c>0$ and real $r$, an exponential $e^{ct}$ eventually dominates $t^r$. Thus, when $p>1$, exponential decay dominates any power of $t$; when $p<1$, exponential growth dominates it.

At $p=1$, the exponential factor disappears, leaving $\int t^{-q}\,dt$. This integral converges at infinity exactly when $q>1$, as follows by integrating $t^{-q}$ for $q\ne1$, and integrating $1/t$ when $q=1$.

For $p\le0$, the magnitude terms do not tend to zero except possibly when $p=0,\ q>0$. In that remaining case the Integral Test applies, and the transformed integral is $\int e^t t^{-q}\,dt$, which diverges. So the magnitude series diverges throughout $p\le0$.

Absolute-convergence classification: The magnitude series converges if $p>1$, for any real $q$, or if $p=1$ and $q>1$. It diverges for $p<1$, and also for $p=1,\ q\le1$.

Putting the two tests together

Ordinary convergence and absolute convergence give the complete classification. In the table, “conditional” means the alternating series converges while its magnitude series diverges.

Parameter conditions	Classification of the alternating series
$p>1$, any $q$	Absolutely convergent
$p=1,\ q>1$	Absolutely convergent
$0<p<1$, any $q$	Conditionally convergent
$p=1,\ q\le1$	Conditionally convergent
$p=0,\ q>0$	Conditionally convergent
$p<0$, any $q$; or $p=0,\ q\le0$	Divergent

The logarithmic threshold at $p=1$ is particularly important: for the magnitude series, $q>1$ gives absolute convergence, while $q\le1$ does not. Alternation can still make the signed series converge in the latter range, provided its terms decrease to zero.

Worked classifications

Worked Example: A conditional series with a power below one

Classify

\sum_{n=2}^{\infty}\frac{(-1)^{n-1}}{\sqrt n\,(\log n)^3}.

Here $p=\tfrac12$ and $q=3$. The term magnitude is $b_n=n^{-1/2}(\log n)^{-3}$. Using the derivative formula,

f'(x)=-\frac{f(x)}{x}\left(\frac12+\frac{3}{\log x}\right)<0 \qquad (x>1).

Thus $b_n$ decreases, and since $p>0$, $b_n\to0$. The Alternating-Series Test proves ordinary convergence. For absolute convergence, $p=\tfrac12<1$; the transformed integral has integrand $e^{t/2}t^{-3}$, which grows without bound eventually and has a divergent integral. The magnitude series diverges. Therefore the alternating series is conditionally convergent.

Check: its terms approach zero and alternate, so the alternating-series conclusion applies; its absolute-value series falls in the $p<1$ divergent range. The two tests agree with the classification.

Worked Example: Absolute convergence at the logarithmic threshold

Classify

\sum_{n=2}^{\infty}\frac{(-1)^{n-1}}{n(\log n)^{3/2}}.

Here $p=1$ and $q=\tfrac32$. For the magnitude function,

f'(x)=-\frac{f(x)}{x}\left(1+\frac{3}{2\log x}\right)<0 \qquad (x>1).

The magnitude terms decrease to zero. To test absolute convergence, substitute $t=\log x$ in the integral:

\int_2^\infty\frac{dx}{x(\log x)^{3/2}} =\int_{\log 2}^\infty t^{-3/2}\,dt =\left[-2t^{-1/2}\right]_{\log 2}^{\infty} =\frac{2}{\sqrt{\log 2}}.

The integral is finite, so the magnitude series converges by the Integral Test. Hence the alternating series converges absolutely. This is stronger than ordinary convergence and does not depend on the signs alternating.

Check: $p=1,\ q=\tfrac32>1$, which is exactly the absolutely convergent boundary case in the classification.

Worked Example: Conditional convergence when the power of n is zero

Classify

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

This has $p=0$ and $q=1$. The magnitude function is $f(x)=1/\log x$, and

f'(x)=-\frac{1}{x(\log x)^2}<0 \qquad (x>1), \qquad \lim_{x\to\infty}\frac{1}{\log x}=0.

The Alternating-Series Test gives ordinary convergence. For absolute convergence, $\log n\le n$ for $n\ge2$, so

\frac{1}{\log n}\ge\frac{1}{n}.

The harmonic series $\sum 1/n$ diverges, so comparison shows that $\sum 1/\log n$ diverges as well. Therefore the alternating series is conditionally convergent.

Check: $p=0,\ q>0$ gives ordinary convergence, but the magnitude series diverges. This matches the classification.

A common mistake

A frequent error is to conclude that every alternating series with $p=1$ converges absolutely. The power $1/n$ alone does not settle absolute convergence when a logarithmic factor is present.

\text{Incorrect: }\quad \sum_{n=2}^{\infty}\frac{1}{n\log n} \text{ converges because it resembles a }p\text{-series with }p=1.

That reasoning ignores the logarithmic factor and misuses the $p$-series criterion. In fact, for $p=1,\ q=1$, substitution $t=\log x$ gives

\int_2^\infty\frac{dx}{x\log x} =\int_{\log 2}^\infty\frac{dt}{t} =\infty.

Thus the magnitude series diverges. The alternating series itself does converge, because its terms decrease to zero, so the correct classification is conditional convergence—not absolute convergence.

Why the distinction matters

The alternating-series test is efficient for deciding ordinary convergence, but it does not decide absolute convergence. In this family, that distinction is visible in the parameter ranges: alternation permits convergence even when the positive magnitude series diverges. Checking both series prevents an incomplete classification.

A reliable procedure is to first test whether the term magnitudes decrease to zero, then classify the magnitude series. At the boundary $p=1$, evaluate the logarithmic exponent separately: it is the value of $q$, not alternation, that determines absolute convergence.

Check Your Understanding

For each question, distinguish ordinary convergence from absolute convergence.

For which values of $p$ and $q$ do the terms of the alternating series fail to approach zero?
Classify the series when $p=\tfrac34$ and $q=-2$.
For $p=1$, what condition on $q$ gives absolute convergence?
Why does $p=0,\ q>0$ give ordinary convergence but not absolute convergence?
Classify the series when $p=2$ and $q=-5$.

Parameter conditions	Classification of the alternating series
\(p>1\), any \(q\)	Absolutely convergent
\(p=1,\ q>1\)	Absolutely convergent
\(0<p<1\), any \(q\)	Conditionally convergent
\(p=1,\ q\le1\)	Conditionally convergent
\(p=0,\ q>0\)	Conditionally convergent
\(p<0\), any \(q\); or \(p=0,\ q\le0\)	Divergent

Logarithmic Factors in Alternating Series

What You'll Learn

Why logarithmic factors need careful attention

First test the alternating series itself

Classifying the absolute-value series

Putting the two tests together

Worked classifications

Worked Example: A conditional series with a power below one

Worked Example: Absolute convergence at the logarithmic threshold

Worked Example: Conditional convergence when the power of n is zero

A common mistake

Why the distinction matters

Check Your Understanding

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