Using Cauchy's integral test, examine the convergence
Using Cauchy's integral test, examine the convergence of: $$ \sum_{n=2}^{\infty} \frac{1}{n(\ln n)^\alpha}, \qquad \alpha>0. $$
Explanation of Method
The function $f(x)=\frac1{x(\ln x)^\alpha}$ is positive, continuous, and decreasing on $[2,\infty)$, so the integral test applies. The substitution $u=\ln x$ converts the integral into $\int_{\ln2}^\infty u^{-\alpha}\,du$ — exactly the same type of integral as the plain $p$-series of Problem 94, just shifted to start at $\ln2$ instead of $1$, reproducing the identical threshold $\alpha>1$.
- Verify $f(x)=\frac1{x(\ln x)^\alpha}$ is decreasing for $x\ge2$.
- Substitute $u=\ln x$ to reduce to a $p$-type integral in $u$.
- Apply the integral test, matching the threshold $\alpha>1$.
Step 1 · Verify $f$ is decreasing
$$ f'(x) = -\frac1{x^2(\ln x)^\alpha}\left(1+\frac\alpha{\ln x}\right) < 0 \quad\text{for } x\ge2, $$
since both terms in the parentheses are positive there. So $f$ is positive, continuous, and decreasing on $[2,\infty)$.
Step 2 · Substitute $u=\ln x$
With $du=\frac1x\,dx$, and $x=2\Rightarrow u=\ln2$:
$$ \int_2^{\infty} \frac{dx}{x(\ln x)^\alpha} = \int_{\ln2}^{\infty} u^{-\alpha}\,du. $$
Step 3 · Evaluate and apply the integral test
For $\alpha\ne1$: $\int_{\ln2}^\infty u^{-\alpha}du = \lim_{t\to\infty}\frac{t^{1-\alpha}-(\ln2)^{1-\alpha}}{1-\alpha}$, which is finite $=\dfrac{(\ln2)^{1-\alpha}}{\alpha-1}$ if $\alpha>1$, and infinite if $\alpha<1$. For $\alpha=1$: $\int_{\ln2}^\infty\frac{du}u=\infty$. So the series converges iff $\alpha>1$.
Summary Table
| Value of $\alpha$ | Conclusion |
|---|---|
| $\alpha>1$ | Converges, value $\dfrac{(\ln2)^{1-\alpha}}{\alpha-1}$ |
| $0<\alpha\le1$ | Diverges |
Additional Notes
- The substitution $u=\ln x$ is the key technique that recurs throughout Problem 96's more elaborate iterated-logarithm series, where a second substitution $v=\ln u$ peels off yet another logarithmic layer.
- This series, with $\alpha=1$, is the classical example showing that $\sum\frac1{n\ln n}$ diverges even though its terms decay faster than $\frac1n$ alone — the extra logarithmic factor slows the decay just barely too little to secure convergence, exactly mirroring why $\sum\frac1n$ itself diverges.
- This result is what makes Cauchy's condensation test (Problem 98) so useful in practice: applying condensation to $a_n=\frac1{n(\ln n)^\alpha}$ produces $2^na_{2^n}=\frac1{(n\ln2)^\alpha}\sim Cn^{-\alpha}$, instantly recovering the same threshold $\alpha>1$ without any integral at all.
Result
The series converges if and only if $\alpha>1$: