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: $$ \sum_{n=3}^{\infty} \frac{1}{(\ln n)^{\ln \ln n}}. $$
Explanation of Method
Rewriting the term as $e^{-(\ln\ln n)^2}$ shows its decay is governed by $(\ln\ln n)^2$ — a quantity that grows far more slowly than $\ln n$ itself. Comparing $a_n$ against $\frac1n$ via the limit comparison test shows the ratio $n\cdot a_n\to\infty$, meaning $a_n$ decays more slowly than the divergent harmonic series, so the series diverges.
- Rewrite the term in exponential form, simplifying the exponent to $-(\ln\ln n)^2$.
- Compare the growth rate of $(\ln\ln n)^2$ against $\ln n$.
- Apply the limit comparison test against $\sum1/n$, showing the ratio diverges to $\infty$.
Step 1 · Rewrite in exponential form
$$ a_n = \frac1{(\ln n)^{\ln\ln n}} = e^{-\ln\ln n\cdot\ln(\ln n)} = e^{-(\ln\ln n)^2}. $$
Step 2 · Compare $(\ln\ln n)^2$ against $\ln n$
Let $m=\ln n$. As $n\to\infty$, $m\to\infty$, and:
$$ \frac{(\ln\ln n)^2}{\ln n} = \frac{(\ln m)^2}{m} \to 0, $$
since $(\ln m)^2$ grows far more slowly than $m$. So $(\ln\ln n)^2$ is eventually much smaller than $\ln n$ (for any fixed proportion), meaning $a_n=e^{-(\ln\ln n)^2}$ decays much more slowly than $e^{-c\ln n}=n^{-c}$ for any fixed $c>0$.
Step 3 · Limit comparison test against $\sum1/n$
$$ \frac{a_n}{1/n} = n\cdot e^{-(\ln\ln n)^2} = e^{\ln n - (\ln\ln n)^2} = e^{m-(\ln m)^2}. $$
Since $m$ grows linearly while $(\ln m)^2$ grows only logarithmically-squared, $m-(\ln m)^2\to\infty$, so:
$$ \lim_{n\to\infty} \frac{a_n}{1/n} = \infty. $$
Since $\sum\frac1n$ diverges and $a_n$ eventually exceeds any constant multiple of $\frac1n$, $\sum_{n=3}^{\infty}\frac1{(\ln n)^{\ln\ln n}}$ diverges too.
Summary Table
| Quantity | Value |
|---|---|
| General term | $a_n=e^{-(\ln\ln n)^2}$ |
| $(\ln\ln n)^2/\ln n$ | $\to0$ |
| $n\cdot a_n$ | $\to\infty$ |
| Conclusion | Diverges |
Additional Notes
- This series diverges "just barely": the terms $a_n$ do tend to $0$, and in fact decay faster than $1/(\ln n)^\alpha$ for any fixed $\alpha$ (making it converge more plausibly than the always-divergent Problem 43 series), but not quite fast enough to beat $1/n$.
- The direct contrast with Problem 46 is instructive: replacing the exponent $\ln n$ (Problem 46, convergent) with the much slower $\ln\ln n$ (this problem, divergent) shows just how delicately the convergence threshold depends on the precise rate at which the exponent itself grows.
- This is a good example of the general principle that whenever the general term's decay rate can be shown to fall strictly below $1/n$ (in the sense that $n\cdot a_n\to\infty$), the comparison test with the harmonic series settles divergence, without needing an exact closed-form asymptotic.
Result
The series diverges: