Why a Logarithmic Factor Changes the Boundary Test
At a power-series boundary, the powers of the endpoint can leave a series whose terms decrease almost as slowly as the harmonic terms \(1/n\). A logarithmic factor can change whether the positive-term series converges, while a sign alternation can change the result at the opposite endpoint.
For a real number \(q\), consider the boundary series obtained from
Here \(\log\) means the natural logarithm. The sum starts at \(n=2\), so \(\log n\) is positive and the denominator is defined for every real \(q\). At \(x=1\), the terms are positive; at \(x=-1\), they alternate in sign. We will determine convergence at these two endpoints, not the radius of convergence or behavior at other points.
The Positive Endpoint: A Threshold at \(q=1\)
At \(x=1\), the question is whether
converges. The threshold is not \(q>0\), but \(q>1\). A useful way to see the threshold is to group the terms in blocks whose sizes double.
For each integer \(k\geq 1\), put the indices \(n=2^k,\ldots,2^{k+1}-1\) in one block. This block contains \(2^k\) terms. Within it, \(2^k\leq n<2^{k+1}\) and \(k\log 2\leq\log n<(k+1)\log 2\). Thus, for each fixed real \(q\), the terms in this block are bounded above and below by positive constant multiples of \(2^{-k}k^{-q}\). The constants can depend on \(q\), but not on \(k\). Multiplying by the \(2^k\) terms shows that the sum of a block is bounded above and below by positive constant multiples of \(k^{-q}\).
Because the terms are positive, the original series converges if and only if its consecutive block sums have a finite total. The comparison therefore reduces the question to the p-series \(\sum 1/k^q\), which converges exactly when \(q>1\). This proves the positive-endpoint result, including divergence when \(q=1\) and when \(q<1\).
Worked Example: The Positive Endpoint When \(q=1\)
Consider \(\sum_{n=2}^{\infty}1/(n\log n)\). In the block \(2^k\leq n<2^{k+1}\), there are \(2^k\) terms. For every term in that block, \(n<2^{k+1}\) and \(\log n<(k+1)\log 2\). Therefore
Adding the \(2^k\) terms gives a block sum greater than \(1/[2(k+1)\log 2]\). Summing these lower bounds over \(k\geq1\) gives a positive constant multiple of the harmonic series \(\sum_{k=1}^{\infty}1/(k+1)\), which diverges. So the positive-endpoint series diverges. The check is direct: its block sums are bounded below by a divergent series, so the original positive series cannot have a finite sum.
Worked Example: The Positive Endpoint When \(q=3/2\)
Now consider \(\sum_{n=2}^{\infty}1/[n(\log n)^{3/2}]\). The block comparison reduces its convergence to
This is a p-series with exponent \(3/2>1\), so it converges. Consequently, the original positive-endpoint series converges. The check is that the block sums are bounded above by a constant multiple of \(k^{-3/2}\), whose sum is finite; hence the total of all the positive blocks is finite as well.
The Negative Endpoint: Alternation for Every Real \(q\)
At \(x=-1\), the boundary series is
Its term magnitudes are \(a_n=1/[n(\log n)^q]\), which are positive for \(n\geq2\). To apply the alternating-series criterion, two facts are needed: \(a_n\) must tend to zero, and it must decrease from some point onward. “From some point onward” matters: a finite number of initial terms do not affect convergence.
Define \(f_q(x)=1/[x(\log x)^q]\) for \(x>1\). Differentiation gives
The factor outside the parentheses is negative. If \(q\geq0\), the parenthetical factor is positive for \(x>1\), so \(f_q\) decreases there. If \(q<0\), that factor is positive whenever \(\log x>-q\), or equivalently \(x>e^{-q}\). Thus, for every real \(q\), the term magnitudes eventually decrease.
They also tend to zero. For \(q\geq0\), once \(\log n\geq1\), we have \(0<a_n\leq1/n\), which tends to zero. For \(q<0\), put \(r=-q>0\). Then \(a_n=(\log n)^r/n\). With \(t=\log n\), this is \(t^r/e^t\), which tends to zero: choose an integer \(m>r\); since \(e^t\geq t^m/m!\) for \(t>0\),
The alternating-series criterion now applies for every real \(q\). Briefly, its convergence mechanism is that consecutive terms cancel in pairs. Once the magnitudes decrease, the paired partial sums move monotonically and remain bounded; the difference between neighboring partial sums is a term whose magnitude tends to zero. The partial sums therefore approach a common limit.
Worked Example: \(q=1/2\) at Both Endpoints
At \(x=1\), the terms are \(1/[n(\log n)^{1/2}]\). Since \(q=1/2\leq1\), the positive-endpoint threshold shows that this series diverges. At \(x=-1\), the terms are \((-1)^n/[n(\log n)^{1/2}]\). The derivative calculation has \(1+q/\log x=1+1/(2\log x)>0\) for \(x>1\), so the magnitudes decrease. Also \(a_n\leq1/n\) once \(\log n\geq1\), and hence \(a_n\to0\). The alternating-series criterion gives convergence at \(x=-1\). The check is that the two endpoints have the same magnitudes but different signs: the positive series diverges by the threshold, whereas the alternating series meets both convergence conditions.
Worked Example: A Negative Exponent, \(q=-2\)
Here the magnitude is \(a_n=(\log n)^2/n\). At \(x=1\), the positive series diverges because \(-2\leq1\). At \(x=-1\), it is important not to assume the magnitudes decrease from the first term. The derivative factor is \(1-2/\log x\), which is positive when \(x>e^2\). Thus the magnitudes decrease for all sufficiently large \(n\). To check their limit, set \(t=\log n\) and use \(e^t\geq t^3/3!\):
The alternating series therefore converges at \(x=-1\). This example verifies that a negative exponent does not prevent alternating convergence: the magnitudes eventually decrease and still tend to zero.
Convergence at a Glance
| Endpoint or test | Result for real \(q\) |
|---|---|
| \(x=1\): positive-term series | Converges if and only if \(q>1\) |
| \(x=-1\): alternating series | Converges for every real \(q\) |
| Absolute values at \(x=-1\) | Converge if and only if \(q>1\) |
When \(q\leq1\), the negative-endpoint series converges while its absolute-value series diverges. This is called conditional convergence. The name records that convergence depends on the alternating signs; it does not mean that the terms fail to tend to zero.
A Common Mistake
A frequent error is to conclude that the alternating series at \(x=-1\) diverges whenever its absolute-value series diverges. For example, at \(q=1\), that mistaken calculation is: “\(\sum 1/(n\log n)\) diverges, so \(\sum (-1)^n/(n\log n)\) diverges.” The conclusion does not follow. The first series tests absolute values; the second has alternating signs. In the second, the magnitudes tend to zero and decrease, so the alternating-series criterion proves convergence.
The correction is to test the two questions separately. First test whether the series of magnitudes converges; that happens exactly when \(q>1\). Then test the alternating series itself; its magnitudes tend to zero and are eventually decreasing for every real \(q\), so it converges for every real \(q\).
Check Your Understanding
Use the endpoint criteria and the conditions for alternating convergence to answer these questions.
- For which real values of \(q\) does \(\sum_{n=2}^{\infty}1/[n(\log n)^q]\) converge?
- At \(q=1\), what is the convergence result at \(x=1\), and what is it at \(x=-1\)?
- Why does the block comparison reduce the positive-endpoint test to a p-series?
- For \(q=-3\), why is the magnitude eventually decreasing, even if it is not decreasing for every \(n\geq2\)?
- For which real \(q\) does the negative-endpoint series converge but fail to converge absolutely?