Using Cauchy's integral test, examine the convergence
Using Cauchy's integral test, examine the convergence of: $$ 1^\circ\; \sum_{n=2}^{\infty} \frac{1}{n \ln^3 n \ln^2(\ln n)}; \qquad 2^\circ\; \sum_{n=2}^{\infty} \frac{1}{n \ln n \ln(\ln n)}. $$
Explanation of Method
Both series call for the double substitution $u=\ln x$ followed by $v=\ln u$. Part $1^\circ$'s extra powers ($\ln^3$ and $\ln^2$) leave an integral dominated by $u^{-3}$, which converges comfortably; Part $2^\circ$'s single powers reduce all the way down to $\int\frac{dv}v$, the divergent harmonic-type integral — the same boundary behavior as $\sum1/n$ itself, just nested one level deeper in logarithms.
- Substitute $u=\ln x$ in each integral to remove the outer factor $1/x$.
- Evaluate Part $1^\circ$'s integral, bounding it by a convergent power of $u$.
- Evaluate Part $2^\circ$'s integral via a second substitution $v=\ln u$, reducing to $\int dv/v=\infty$.
Step 1 · Part $1^\circ$: substitute and bound
With $u=\ln x$:
$$ \int_3^{\infty} \frac{dx}{x\ln^3x\ln^2(\ln x)} = \int_{\ln3}^{\infty} \frac{du}{u^3\ln^2u}. $$
Since $\ln^2u\ge1$ eventually (once $u>e$), $\dfrac1{u^3\ln^2u}\le\dfrac1{u^3}$ there, and $\int_{\ln3}^\infty u^{-3}du=\dfrac1{2(\ln3)^2}<\infty$: the integral converges by comparison, so by the integral test Part $1^\circ$ converges.
Step 2 · Part $2^\circ$: double substitution
With $u=\ln x$, then $v=\ln u$:
$$ \int_3^{\infty} \frac{dx}{x\ln x\ln(\ln x)} = \int_{\ln3}^{\infty} \frac{du}{u\ln u} = \int_{\ln(\ln3)}^{\infty} \frac{dv}{v} = \infty. $$
Step 3 · Conclude
By the integral test: Part $1^\circ$ converges (integral bounded by a convergent $u^{-3}$-type integral); Part $2^\circ$ diverges (integral reduces exactly to $\int dv/v=\infty$).
Summary Table
| Part | Reduced integral | Conclusion |
|---|---|---|
| $1^\circ$ | $\int_{\ln3}^\infty u^{-3}\ln^{-2}u\,du$ (converges) | Converges |
| $2^\circ$ | $\int_{\ln(\ln3)}^\infty v^{-1}dv$ | Diverges |
Additional Notes
- These two series form a natural bracketing pair: any exponent combination whose reduced integral is dominated by $u^{-p}$ with $p>1$ (as in Part $1^\circ$) converges, while reducing all the way to the bare $\int dv/v$ form (as in Part $2^\circ$) always diverges — this is the logarithmic analogue of the plain $p$-series threshold from Problem 94.
- The general pattern $\sum\frac1{n\ln n\ln(\ln n)\cdots(\ln^{(k)}n)^\alpha}$ converges iff the very last (innermost) exponent $\alpha>1$, with all outer logarithmic layers required to have exponent exactly $1$ to allow the substitution chain to fully unwind — Part $2^\circ$ is exactly this pattern with $k=2$ and $\alpha=1$, landing right on the divergent boundary.
- Part $1^\circ$'s convergence doesn't actually need the full strength of $\ln^3n\ln^2(\ln n)$: even $\ln^{1+\varepsilon}n$ alone (any exponent above $1$ on the first logarithm) would suffice by the same style of argument as Problem 95, since the extra $\ln^2(\ln n)$ factor here is not essential to convergence, only to the specific bound used.
Result
Part $1^\circ$ converges; Part $2^\circ$ diverges: