For each \((a,b,c)\in\RR^3\), consider the series
\[ \sum_{n=3}^{\infty}\frac{a^n}{n^b(\log n)^c}. \]
Determine the values of \((a,b,c)\) for which the series
converges absolutely;
converges but not absolutely;
diverges.
Solution
Write \[ u_n=\frac{a^n}{n^b(\log n)^c}. \]
1. If \(\abs{a}<1\), the series converges absolutely; if \(\abs{a}>1\), itdiverges.
The \(n\)th root of \(\abs{u_n}\) satisfies \[ \abs{u_n}^{1/n} =\abs{a}\,n^{-b/n}(\log n)^{-c/n} \longrightarrow \abs{a}. \] The root test gives absolute convergence when \(\abs{a}<1\) and divergence when \(\abs{a}>1\).
2. For \(\abs{a}=1\), the series\[ \sum_{n=3}^\infty \abs{u_n}=\sum_{n=3}^\infty \frac1{n^b(\log n)^c} \]converges exactly when \(b>1\), or when \(b=1\) and \(c>1\).
If \(b=1\), the integral test compares the series with \[ \int_3^\infty \frac{dx}{x(\log x)^c}. \] With \(u=\log x\), this becomes \[ \int_{\log 3}^\infty u^{-c}\,du, \] which converges exactly when \(c>1\).
If \(b>1\) and \(c\ge0\), the summand is eventually bounded by \(n^{-b}\). If \(b>1\) and \(c<0\), choose \(0<\varepsilon<b-1\). Since \((\log n)^{\abs{c}}=o(n^\varepsilon)\), the summand is eventually bounded by \(n^{-(b-\varepsilon)}\), and \(b-\varepsilon>1\). In both cases the series converges.
If \(b<1\), choose \(0<\varepsilon<1-b\). When \(c>0\), the estimate \((\log n)^c=o(n^\varepsilon)\) gives, for large \(n\), \[ \frac1{n^b(\log n)^c}\ge \frac1{n^{b+\varepsilon}}, \] and \(b+\varepsilon<1\), so the series diverges. When \(c\le0\), the summand is at least \(n^{-b}\) for \(n\ge3\), and the series diverges because \(b<1\).
3. For \(a=1\), the series converges exactly when it convergesabsolutely.
For \(a=1\) every term is positive, so \(u_n=\abs{u_n}\).
4. For \(a=-1\), the series converges exactly when\[ b>0 \qquad\text{or}\qquad b=0,\ c>0. \]
Write \[ d_n=\frac1{n^b(\log n)^c}>0, \] so that \(u_n=(-1)^nd_n\). The sequence \(d_n\) tends to \(0\) exactly when \(b>0\), or \(b=0\) and \(c>0\); otherwise \(d_n\not\to0\) and the series diverges. In the two convergent cases, \[ \frac{d}{dx}\log\frac1{x^b(\log x)^c} =-\frac1x\left(b+\frac{c}{\log x}\right)<0 \] for large \(x\), so \(d_n\) is eventually decreasing, and the alternating series test gives convergence.
5. The series\[ \boxed{ \begin{aligned} &\text{converges absolutely iff } \abs{a}<1, \text{ or } \abs{a}=1 \text{ and } (b>1 \text{ or } b=1,\ c>1);\\ &\text{converges but not absolutely iff } a=-1 \text{ and } (0<b<1,\ \text{or } b=1,\ c\le1,\ \text{or } b=0,\ c>0);\\ &\text{diverges in all other cases.} \end{aligned} } \]
Steps 1 and 2 give the absolute-convergence region. Conditional convergence requires convergence without absolute convergence; by step 1 this forces \(\abs{a}=1\), by step 3 it forces \(a=-1\), and step 4 together with step 2 gives the listed parameters. Every other parameter value gives a divergent series by steps <1>1–<1>4.
6. Q.E.D.
Step 5 answers all three parts.