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: $$ \sum_{n=3}^{\infty} \frac{1}{(\ln n)^{\ln n}}. $$
Explanation of Method
Writing the term as $e^{-\ln n\cdot\ln(\ln n)}$ shows that once $\ln(\ln n)$ exceeds any fixed constant $c$ — which it eventually does, since $\ln n\to\infty$ — the denominator $(\ln n)^{\ln n}$ overtakes $n^c$. Taking $c=2$ dominates the series by the convergent $\sum1/n^2$, and in fact the denominator eventually beats every polynomial power of $n$.
- Rewrite the term in exponential form.
- Show that once $\ln(\ln n)\ge c$, the denominator satisfies $(\ln n)^{\ln n}\ge n^c$.
- Apply the comparison test using $c=2$ against $\sum1/n^2$.
Step 1 · Rewrite in exponential form
$$ a_n = \frac1{(\ln n)^{\ln n}} = e^{-\ln n\cdot\ln(\ln n)}. $$
Since $\ln n\to\infty$ and $\ln(\ln n)\to\infty$ (more slowly), the exponent $\to-\infty$, so $a_n\to0$.
Step 2 · Bound the denominator using a fixed constant $c$
For any fixed $c>0$, once $n$ is large enough that $\ln(\ln n)\ge c$ (e.g. $c=2$ requires $n\ge e^{e^2}\approx1618$):
$$ (\ln n)^{\ln n} = e^{\ln n\cdot\ln(\ln n)} \ge e^{c\ln n} = n^c. $$
Step 3 · Apply the comparison test with $c=2$
For sufficiently large $n$: $\dfrac1{(\ln n)^{\ln n}} \le \dfrac1{n^2}$. Since $\sum\frac1{n^2}$ converges, by the comparison test $\sum_{n=3}^{\infty}\frac1{(\ln n)^{\ln n}}$ converges too — and since $c$ may be taken arbitrarily large, the terms in fact decay faster than any fixed polynomial rate $1/n^c$.
Summary Table
| Quantity | Value |
|---|---|
| General term | $a_n=e^{-\ln n\cdot\ln(\ln n)}$ |
| Eventual bound (with $c=2$) | $a_n\le1/n^2$ for $n\ge e^{e^2}$ |
| Conclusion | Converges (super-polynomially fast) |
Additional Notes
- Numerically, $(\ln n)^{\ln n}$ eventually overtakes $n^2$: at $n=1000$, $(\ln n)^{\ln n}\approx5\times10^5$ versus $n^2=10^6$ (not yet crossed), but by $n=e^{10}\approx22000$, $(\ln n)^{\ln n}\approx10^{10}$ versus $n^2\approx5\times10^8$ — well past the crossover.
- The root test applied to this series gives an inconclusive limit of $1$ (since $\frac{\ln n\cdot\ln(\ln n)}n\to0$), which is why a direct comparison bound — rather than a standard convergence test — is needed here.
- This problem contrasts with Problem 47's $\sum1/(\ln n)^{\ln\ln n}$: replacing the exponent $\ln n$ with the much slower-growing $\ln\ln n$ flips the outcome from convergent to divergent, showing how sensitive these iterated-logarithm series are to the exact rate of exponent growth.
Result
The series converges: