For which pairs of real numbers \((a,b)\) does the series \(\sum_{n=3}^{\infty}n^a(\log n)^b\) converge?
Solution
1. For every \(c>0\) and \(\varepsilon>0\),\[ (\log n)^c=o(n^\varepsilon). \]
Taking logarithms, it is enough to show \[ c\log\log n-\varepsilon\log n\longrightarrow-\infty. \] But \[ \frac{\log\log n}{\log n}\longrightarrow0, \] since with \(u=\log n\) this is \(\log u/u\to0\), for example by l’Hospital’s rule. Hence the negative term \(-\varepsilon\log n\) dominates.
2. If \(a<-1\), then the series converges for every real \(b\).
If \(b\le0\), then \((\log n)^b\le1\) for \(n\ge3\), so \[ n^a(\log n)^b\le n^a, \] and \(\sum n^a\) converges because \(a<-1\).
If \(b>0\), choose \(\varepsilon>0\) with \[ a+\varepsilon<-1. \] By step 1, for all sufficiently large \(n\), \[ (\log n)^b\le n^\varepsilon. \] Hence \[ n^a(\log n)^b \le n^{a+\varepsilon}, \] and the comparison series converges.
3. If \(a>-1\), then the series diverges for every real \(b\).
If \(b\ge0\), then \((\log n)^b\ge1\) for \(n\ge3\), so \[ n^a(\log n)^b\ge n^a. \] The series \(\sum n^a\) diverges because \(a>-1\).
If \(b<0\), choose \(\varepsilon>0\) with \[ a-\varepsilon>-1. \] Applying step 1 with \(c=-b>0\) gives, for all sufficiently large \(n\), \[ (\log n)^{-b}\le n^\varepsilon. \] Taking reciprocals, \[ (\log n)^b\ge n^{-\varepsilon}, \] and therefore \[ n^a(\log n)^b \ge n^{a-\varepsilon}. \] The series on the right diverges.
4. If \(a=-1\), then the series converges exactly when \(b<-1\).
Consider \[ f(x)=\frac{(\log x)^b}{x}. \] For all sufficiently large \(x\) this function is positive and decreasing, because \[ f'(x) = x^{-2}(\log x)^{b-1}(b-\log x). \] Thus the integral test applies. With \(u=\log x\), \[ \int_3^\infty\frac{(\log x)^b}{x}\,dx = \int_{\log3}^\infty u^b\,du, \] which converges exactly when \(b<-1\).
5. Therefore the convergence region is\[ \boxed{ \{(a,b)\in\RR^2:a<-1\} \;\cup\; \{(-1,b):b<-1\}. } \]
Steps <1>2–<1>4 cover the three mutually exclusive cases \(a<-1\), \(a=-1\), and \(a>-1\).
6. Q.E.D.
Step 5 gives the complete classification.