# Convergence of $\sum_{x=2}^{\infty}\left( \frac{\ln\left(\ln(x)\right)}{x^2}\right)$

I want to prove that the infinite series defines by:

$$\sum_{x=2}^{\infty} \left(\frac{\ln\left(\ln(x)\right)}{x^2}\right)$$

converges.

I have shown that $$\lim_{x\to\infty} \left( \frac{\ln\left(\ln(x)\right)}{x^2} \right) = 0$$

But proving that $$\frac{\ln\left(\ln(x)\right)}{x^2}$$ is monotonically decreasing appears too complicated.

Also i tried : $$\frac{\ln\left(\ln(x)\right)}{x^2} = \frac{\ln\left(\circ(x)\right)}{x^2} = \frac{\circ\left(\circ(x)\right)}{x^2} = \circ\left(\frac{1}{x}\right)$$ but there I know that the series converges so there has to be something wrong with my reasoning.

Any ideas?

(I consider $$x$$ to be of integer values, probably should have replaced with $$n$$ or $$k$$, but I think it's the same.)

• For large $x$ we have $\ln x\lt x^{1/2}$. Iterating this, we have $\ln(\ln x)\lt\ln(x^{1/2})={1\over2}\ln x\lt{1\over2}x^{1/2}$. Commented Feb 21, 2018 at 16:08
• Wouldn't you need $x$ starting at 2? Commented Feb 21, 2018 at 16:08
• One minor error: your function is not a number in $x=0$ and $x=1$ unless you entered the sequence incorrectly. Commented Feb 21, 2018 at 16:10
• Yeah $x$ starts at $2$, force of of habit, i will edit that. @BarryCipra could you give a more explicit solution. I didn't exactly get what you are getting at. Commented Feb 21, 2018 at 16:11
• @IliasKoutroumpas, for large $x$, we wind up with $\sum{\ln(\ln x)\over x^2}\lt\sum{1\over x^{3/2}}$. Since $3/2\gt1$, the $p$-test tells us $\sum{1\over x^{3/2}}$ converges. Commented Feb 21, 2018 at 16:14

Use Ermakoff' test: $$\lim_\limits{x\to\infty} \frac{e^xf(e^x)}{f(x)}=k<1 \Rightarrow \text{converges}.$$ The given series converges because: $$\lim_\limits{x\to\infty} \frac{x^2 \cdot \ln x}{e^{x}\cdot \ln{(\ln x)}}=0.$$
Using the Cauchy Condensation test, the original sum converges iff the following series: $$\sum_{k=2}^\infty\frac{\ln(2\ln k)}{2^k}=\sum_{k=2}^\infty\frac{\ln2+\ln\ln k}{2^k}=\frac12\ln2+\sum_{k=2}^\infty\frac{\ln\ln k}{2^k}$$converges. The remaining sum $$\sum_{k=2}^\infty\frac{\ln\ln k}{2^k}$$is smaller than $$\zeta'(2)$$ and so the original sum converges.