Knowing that sum from n=1 to infinity 1/(n^()alpha)
Knowing that $\sum_{n=1}^{\infty}\frac{1}{n^\alpha}$ converges for $\alpha>1$ and diverges for $\alpha\le1$, determine the convergence of: $$ \sum_{n=2}^{\infty} \frac{1}{(\ln n)^\alpha}. $$
Explanation of Method
Since $\ln n$ grows dramatically slower than $n$ or any power $n^\epsilon$, the terms $1/(\ln n)^\alpha$ shrink far too slowly for convergence, no matter how large $\alpha$ is chosen. Both the integral test (via the substitution $u=\ln x$, which turns the integral into one with an exponentially growing integrand) and the Cauchy condensation test confirm divergence for every real $\alpha$.
- Apply the integral test, substituting $u=\ln x$.
- Show the resulting integral diverges since $e^u$ dominates any power $u^\alpha$.
- Confirm independently via the Cauchy condensation test.
Step 1 · Integral test with substitution
Consider $\int_2^{\infty}\frac1{(\ln x)^\alpha}\,dx$. Substituting $u=\ln x$, $dx=e^u\,du$:
$$ \int_2^{\infty}\frac1{(\ln x)^\alpha}\,dx = \int_{\ln2}^{\infty}\frac{e^u}{u^\alpha}\,du. $$
Step 2 · Show the integral diverges
As $u\to\infty$, the exponential $e^u$ grows faster than any polynomial power $u^\alpha$ in the denominator, so $\frac{e^u}{u^\alpha}\to\infty$; the integrand does not even tend to $0$, so the integral diverges for every $\alpha$. By the integral test, $\sum_{n=2}^{\infty}\frac1{(\ln n)^\alpha}$ diverges for every $\alpha$.
Step 3 · Confirm via Cauchy's condensation test
For the decreasing sequence $a_n=\frac1{(\ln n)^\alpha}$, the condensed series is $\sum_k 2^k a_{2^k}$. Since $a_{2^k}=\frac1{(k\ln2)^\alpha}$:
$$ \sum_{k=1}^{\infty} 2^k a_{2^k} = \frac1{(\ln2)^\alpha}\sum_{k=1}^{\infty}\frac{2^k}{k^\alpha}. $$
The exponentially growing numerator $2^k$ dominates the polynomial denominator $k^\alpha$ for any fixed $\alpha$, so this series diverges. By the condensation test, the original series diverges too.
Summary Table
| Method | Result |
|---|---|
| Integral test | $\int_{\ln2}^{\infty}\frac{e^u}{u^\alpha}du=\infty$ |
| Condensation test | $\sum\frac{2^k}{k^\alpha}=\infty$ |
| Conclusion | Diverges for every $\alpha\in\mathbb R$ |
Additional Notes
- This result is qualitatively different from ordinary $p$-series behavior: no exponent $\alpha$, however large, can rescue this series, because $\ln n$ is asymptotically smaller than every positive power of $n$, so $1/(\ln n)^\alpha$ is asymptotically larger than $1/n^\epsilon$ for any $\epsilon>0$.
- The Cauchy condensation test is especially well-suited to slowly-varying sequences like $\ln n$: condensing at powers of $2$ converts logarithmic growth into linear growth ($\ln 2^k = k\ln2$), turning the analysis into a familiar comparison between exponential and polynomial growth rates.
- This divergence, despite the terms $\to0$, echoes the pattern seen with the harmonic series and with $\sum\sin(x/n)$ — a reminder that $a_n\to0$ is necessary but never sufficient for convergence.
Result
The series diverges for every real $\alpha$: