$\bigcup\limits_{1\leq r < s} l^r \neq l^s$

Question

We had to prove for $s > 1$ :

$$\bigcup\limits_{1\leq r < s} l^r \neq l^s$$

, i. e. we have to find an $x \in l^s$ which is for all $r < s$ not in $l^r$.

The solution used the sequence $x_n := (n \log(n)\log(n))^{-1/s}$. It is said that $x \in l^s$ (using integral criterion), but this is not the case: Due to WolframAlpha, we get: $$\int\limits_0^\infty (x \log(x)\log(x))^{-1} dx = \left[-\frac{1}{log(x)}\right]_{x \to 0}^{x\to\infty}$$

But the right side diverges according to WolframAlpha and my own calculations. Which other easy example does in which way prove the given result?

Answer 1

Take $x_n=(n log (n) log (n))^{-1/s}$ for $n \geq 2$ and $x_1=x_2=0$. Then everything works fine. The integral is now from $2$ to $\infty$.