For which values of \(\alpha,\beta\) does \[ \sum_{n=3}^\infty\frac{1}{n^\alpha(\log n)^\beta} \] converge?
Solution
Put \[ a_n\coloneqq\frac{1}{n^\alpha(\log n)^\beta}. \]
1. If \(\alpha>0\), then \((a_n)\) is eventually decreasing. If\(\alpha=0\) and \(\beta>0\), it is decreasing for \(n\geq3\).
For \[ h(x)=x^{-\alpha}(\log x)^{-\beta}, \] one has \[ \frac{h'(x)}{h(x)} = -\frac1x \left( \alpha+\frac{\beta}{\log x} \right). \] If \(\alpha>0\), the quantity in parentheses is positive for all sufficiently large \(x\). If \(\alpha=0\) and \(\beta>0\), it is positive for every \(x>1\). Thus \(h\) is decreasing in the stated ranges.
2. Whenever step 1 applies, Cauchy’s condensation test reduces theseries, up to a positive constant factor, to\[ \sum_{k\geq K} \frac{2^{(1-\alpha)k}}{k^\beta} \]for some sufficiently large integer \(K\).
For sufficiently large \(k\), \[ 2^k a_{2^k} = \frac{2^k} {(2^k)^\alpha(\log 2^k)^\beta} = (\log2)^{-\beta} \frac{2^{(1-\alpha)k}}{k^\beta}. \] Cauchy’s condensation test applies to the positive eventually decreasing sequence after discarding finitely many initial terms.
3. If \(\alpha>1\), the original series converges for every\(\beta\in\RR\).
In step 2 the factor \(2^{(1-\alpha)k}\) decays geometrically. A geometric decay dominates the fixed power \(k^{-\beta}\), so the condensed series converges. Hence the original series converges.
4. If \(0<\alpha<1\), the original series diverges for every\(\beta\in\RR\).
The terms of the condensed series from step 2 are \[ \frac{2^{(1-\alpha)k}}{k^\beta}. \] Since \(1-\alpha>0\), these terms do not tend to zero. Thus the condensed series, and hence the original series, diverges.
5. If \(\alpha=1\), the original series converges exactly when\(\beta>1\).
For \(\alpha=1\), step 2 gives, up to a positive constant factor, \[ \sum_{k\geq K}\frac1{k^\beta}. \] This \(p\)-series converges exactly when \(\beta>1\).
6. If \(\alpha\leq0\), the original series diverges for every\(\beta\in\RR\).
If \(\alpha=0\) and \(\beta>0\), step 2 gives condensed terms \[ \frac{2^k}{k^\beta}, \] which do not tend to zero. If \(\alpha=0\) and \(\beta\leq0\), then \(a_n\) itself does not tend to zero.
If \(\alpha<0\), then \[ a_n = \frac{n^{-\alpha}}{(\log n)^\beta} \] does not tend to zero, because every positive power of \(n\) dominates every fixed power of \(\log n\). Thus the series diverges in all cases with \(\alpha\leq0\).
7. Therefore the series converges exactly for\[ \boxed{ \alpha>1 \quad\text{or}\quad \alpha=1\text{ and }\beta>1 }. \]
Steps <1>3–<1>6 cover all real values of \(\alpha\) and \(\beta\).
8. Q.E.D.
Step 7 is the required classification.