TITLE: Does $\displaystyle\sum x^{\lfloor \ln(n) \rfloor}$ converges ? For $x\in(-1,-\frac{1}e]$
QUESTION [4 upvotes]: Does $\displaystyle\sum x^{\lfloor \ln(n) \rfloor}$ converges?
Let $u_n= x^{\lfloor \ln(n) \rfloor}$

For $|x|\geq 1$, $|u_n|\geq 1$. Then, the serie diverges.
For $x=0$, we have $u_n=0$ for $n\geq 3$, thus the serie converges.
For $x\in (0,1)$, we have, for all $n$ integer :
$$
x^{\ln(n)}\leq x^{\lfloor \ln(n) \rfloor}\leq x^{\ln(n-1)}
$$
As $\ln(x^{\ln(n)})=\ln(n)\ln(x)=\ln(n^{\ln(x)})$, we get
$$
n^{\ln(x)}\leq u_n\leq \frac{n^{\ln(x)}}{x}
$$
Thus, $u_n=O(n^{\ln(x)})$ and $n^{\ln(x)}=O(u_n)$

Therefore by theorem, $\displaystyle\sum u_n$ and $\displaystyle\sum n^{\ln(x)}$ have the same nature of convergence.
Plus, by Riemann series $\displaystyle\sum u_n$ converges if and only if $\ln(x)<-1$
ie. if and only if $x<\frac{1}e$

For $x\in (-\frac{1}e,0)$, we have $|u_n|=|x|^{\lfloor \ln(n) \rfloor}$, thus the serie converge absolutely.


For $x\in(-1,-\frac{1}e]$ I feel the serie does not converges but I was not managed to prove it. 
Any ideas ?
Thank you in advance,

REPLY [1 votes]: Some observations: For $x \in (-1, -{1 \over e})$ you have
$$|x|^{\ln n} \leq  |x^{\lfloor \ln n \rfloor}| \leq |x|^{\ln n - 1} = {1 \over |x|} |x|^{\ln n}$$
$$|x|^{\ln n} = n^{\ln|x|}$$
So since $|\ln |x|| < 1$ here, the series does not converge absolutely. There's still the chance it converges conditionally. But notice $\lfloor \ln n \rfloor$ only switches signs at  $n = \lfloor e^k \rfloor + 1$ for $k$ an integer. 
So between each $\lfloor e^k \rfloor + 1$ and $\lfloor e^{k+1} \rfloor + 1$ you will be adding terms of magnitude within a factor of $|x|$ of $n^{\ln|x|}$, each of which has the same sign. The sum of $n^{\ln|x|}$ over such exponential blocks grows in $k$, so the series will not converge.
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