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

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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.)

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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: $$\lim_\limits{x\to\infty} \frac{x^2 \cdot \ln x}{e^{x}\cdot \ln{(\ln x)}}=0.$$