studying the series $\sum_\limits{n=1}^\infty \frac{1}{\log x^{\log n}}$.

Question

I am studying the series $\displaystyle\sum_\limits{n=1}^\infty \frac{1}{\log x^{\log n}}$.

The series has got a sense $\forall n \ge 1 , n \in \mathbb{N}$ and $\forall x \in (0,1) \cup (1, + \infty)$.

If the series converges then $\left|\sum_\limits{n=1}^\infty \frac{1}{\log x^{\log n}}\right|$ converges.

$$\left|\sum_\limits{n=1}^\infty \frac{1}{\log x^{\log n}}\right| \le \sum_\limits{n=1}^\infty \left|\frac{1}{\log x^{\log n}}\right| = \sum_\limits{n=1}^\infty \frac{1}{\left|\log x^{\log n}\right|} =\sum_\limits{n=1}^\infty \frac{1}{\log x^{\left|\log n\right|}} = \sum_\limits{n=1}^\infty \left|\frac{1}{\log x}\right|^{\log n}.$$

Then we can distinguish three cases:

1. Let $|\frac{1}{\log x}|<1 \iff |\log x|>1 \implies x \in (0, \frac{1}{e}) \cup (e, +\infty)$ . Then $\sum_\limits{n=1}^\infty \left|\frac{1}{\log x}\right|^{\log n}$ diverges $ \implies \left|\sum_\limits{n=1}^\infty \frac{1}{\log x^{\log n}}\right|$ converges.

2. Let $\left|\frac{1}{\log x}\right|>1 \iff |\log x|<1 \iff x \in (\frac{1}{e}, 1) \cup (1,e)$ . Then $\sum_\limits{n=1}^\infty |\frac{1}{\log x}|^{\log n} $ diverges $\implies \left|\sum_\limits{n=1}^\infty \frac{1}{\log x^{\log n}}\right|$ diverges.$

3. Let $|\frac{1}{\log x}|=1 \iff |\log x|=1 \iff x=e \lor x= \frac{1}{e}$ . Then $\sum_\limits{n=1}^\infty\left|\frac{1}{\log x}\right|^{\log n} =\sum_\limits{n=1}^\infty 1$ diverges .

Is it right?

Answer 1

You have to revise your work (especially your first case). Note that the term of the series is defined only for $x>1$. Moreover $$(\log x)^{\log n}=\exp(\log(n)\log (\log(x)))=n^{\log(\log(x)}$$ Hence the series is convergent iff $\log(\log(x))>1$, that is for $x>e^e\approx 15.154$.