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: $$ \sum_{n=3}^{\infty} \frac{1}{(\ln\ln n)^{\ln n}}. $$
Explanation of Method
Rewriting the term as $e^{-\ln n\cdot\ln(\ln\ln n)}$ makes clear that once $\ln\ln n$ exceeds any fixed constant $c$, the denominator $(\ln\ln n)^{\ln n}$ exceeds $n^c$ — so the whole term becomes smaller than $n^{-c}$. Choosing $c>1$ (e.g. $c=2$) shows the series is eventually dominated by a convergent $p$-series, since $\ln\ln n\to\infty$ (however slowly).
- Rewrite the term in exponential form to expose the exponent's structure.
- Show that once $\ln\ln n$ exceeds a fixed constant $c$, $(\ln\ln n)^{\ln n} \ge n^c$.
- Apply the comparison test using $c=2$ against the convergent series $\sum 1/n^2$.
Step 1 · Rewrite in exponential form
$$ a_n = \frac1{(\ln\ln n)^{\ln n}} = e^{-\ln n\cdot\ln(\ln\ln n)}. $$
Since $\ln\ln n\to\infty$ as $n\to\infty$ (however slowly), $\ln(\ln\ln n)\to\infty$ too, so the exponent $\to-\infty$ and $a_n\to0$.
Step 2 · Bound the denominator once $\ln\ln n$ exceeds a constant
If $\ln(\ln\ln n)\ge c$ for some constant $c>0$ (true for all sufficiently large $n$, since $\ln\ln n\to\infty$), then:
$$ (\ln\ln n)^{\ln n} = e^{\ln n\cdot\ln(\ln\ln n)} \ge e^{c\ln n} = n^c. $$
Step 3 · Apply the comparison test with $c=2$
Take $c=2$: for sufficiently large $n$ (specifically once $\ln\ln n\ge e^2$, i.e. $n\ge e^{e^{e^2}}$), $(\ln\ln n)^{\ln n}\ge n^2$, so:
$$ \frac1{(\ln\ln n)^{\ln n}} \le \frac1{n^2}. $$
Since $\sum\frac1{n^2}$ converges, by the comparison test $\sum_{n=3}^{\infty}\frac1{(\ln\ln n)^{\ln n}}$ converges too.
Summary Table
| Quantity | Value |
|---|---|
| General term | $a_n=e^{-\ln n\cdot\ln(\ln\ln n)}$ |
| Eventual bound (with $c=2$) | $a_n\le1/n^2$ |
| Conclusion | Converges |
Additional Notes
- The threshold at which $\ln\ln n$ first exceeds $e^2\approx7.39$ is astronomically large ($n\approx e^{e^{e^2}}$), so numerically the terms of this series decrease extremely slowly at first — but the comparison argument is about eventual, asymptotic behavior, which is all that matters for convergence.
- Since $c$ can be taken arbitrarily large (as $\ln\ln n\to\infty$), the terms $a_n$ eventually decay faster than $1/n^c$ for every fixed $c$ — the series converges "super-polynomially" fast, in stark contrast to Problem 43's $\sum1/(\ln n)^\alpha$, which diverges no matter how large $\alpha$ is.
- This problem and Problem 46 ($\sum1/(\ln n)^{\ln n}$) share the same technique: whenever the base of the exponent grows without bound (here $\ln\ln n$; there $\ln n$), the same comparison-with-any-fixed-power argument applies and yields convergence.
Result
The series converges: