When does $\sum_{n=1}^\infty \frac{\ln(1+nx)}{nx^n}$ converge? I want to know for which values this series 
$$\sum_{n=1}^\infty \frac{\ln(1+nx)}{n x^n}$$ converges.
This series is defined for:
$1+nx>0 \Rightarrow x>-\frac{1}{n}$ which tends to $0 \Rightarrow x \ge 0$.
$nx^n \ne 0 \Rightarrow x^n \ne 0 \Rightarrow x\ne 0 $.
So the series is defined for $x>0$
Now we have to see for which values $x>0$ 
$\displaystyle \lim_{n\rightarrow + \infty} \frac{\ln(1+nx)}{nx^n}= 0$
For $x=0$ we have $nx \sim +\infty \Rightarrow 1+nx \sim nx \Rightarrow  \lim_{n\rightarrow + \infty} \frac{\ln(1+nx)}{nx}= \lim_{n\rightarrow + \infty} \frac{\ln(nx)}{nx}=0$
So it should be
$\displaystyle \lim_{n\rightarrow + \infty} \frac{1}{x ^{n-1}}=0 \Rightarrow \lim_{n\rightarrow + \infty} x ^{n-1}= \infty  \Rightarrow |x|>1$.
This last condition is observed is $x>1$.
Applying the root criterion we have:
$\sqrt[n]{|\frac{\ln(1+nx)}{nx^n}|}=\sqrt[n]{\frac{\ln(1+nx)}{n x^n}}=\frac{\sqrt[n]{\ln(1+nx)}}{x} \sim \frac{\sqrt[n]{\ln(nx)}}{x}$
And now I don't know how to proceed.
Is it right until now? And could someone help me to finish the exercise?
 A: You can apply the ratio test:
If $$a_n=\frac{\ln(1+nx)}{n x^n}$$ then $$\frac{a_{n+1}}{a_{n}}=\color{orange}{\frac{\ln(1+nx+x)}{\ln(1+nx)}}\cdot\color{blue}{\frac{n}{n+1}}\cdot x\xrightarrow{n\to\infty}\color{orange}1\cdot\color{blue}1\cdot x=x$$ so the series diverges if $0<x<1$ and converges if $x>1$.
If $x=1$ then the series can be bounded below by the Harmonic series which diverges.
So the series converges if and only if $x>1$.
A: Let $ x $ be a real from $ \mathbb{R}_{+}^{*} $, and let's denote $ f_{n}:x\mapsto\frac{\ln{\left(1+nx\right)}}{n x^{n}} \cdot $
If $ x<1 $, we have that $ f_{n}\left(x\right)\underset{n\to +\infty}{\longrightarrow}+\infty $, which means the series diverges.
If $ x>1 $, we have that $ f_{n}\left(x\right)=\underset{\overset{n\to +\infty}{}}{\mathrm{o}}\left(x^{-n}\right) $, because $ \lim\limits_{n\to +\infty}{\frac{\ln{\left(1+nx\right)}}{n}}=0 $, and since $ \sum\limits_{n\geq 1}{x^{-n}} $ converges, we get that $ \sum\limits_{n\geq 1}{f_{n}\left(x\right)} $ converges.
If $ x=1 $, since we have $ \left(\forall n\in\mathbb{N}^{*}\right),\ \frac{1}{n}\leq\frac{\ln{\left(1+n\right)}}{n} $, and we know that the harmonic series diverges, we get that $ \sum\limits_{n\geq 1}{\frac{\ln{\left(1+n\right)}}{n}} $ diverges.
