Convergence of $\sum_{n=1}^\infty \ln {\sqrt[n]{1+ \frac{x}{n}}}$ I want to know for which values of $x$ this series 
$$\sum_{n=1}^\infty \ln {\sqrt[n]{1+ \frac{x}{n}}}$$ converges.
This series is defined for:
$\sqrt[n]{1+ \frac{x}{n}}>0 \Rightarrow 1+ \frac{x}{n}>0 \Rightarrow \frac{x}{n}>-1 \Rightarrow x>-n \Rightarrow x>-1$ .
So the series is defined for $x>-1$
$a_n=\ln {\sqrt[n]{1+ \frac{x}{n}}} = \frac{\ln (1+ \frac{x}{n})}{n} \sim 0$
because the logarithmic function grows slower than any power of n.
Applying the root criterion we have:
$\sqrt[n]{|\frac{\ln (1+ \frac{x}{n})}{n}|}=\sqrt[n]{\frac{|\ln (1+ \frac{x}{n})|}{n}} \sim \sqrt[n]{|\ln (1+ \frac{x}{n})|} \sim \sqrt[n]{|\frac{x}{n}|} \sim \sqrt[n]{|x|}<1$ for $|x|<1$
So the given series converges for $-1<x<+1$
In my book the suggested result is $x>-1$ but I don't know why
 A: Note that
$$\sqrt[n]{|x|}=x^{1/n}\to x^0=1$$
so that the root criterion is inconclusive. One can more simply use the limit comparison test to notice essentially what you already had though:
$$\frac1n\ln\left(1+\frac xn\right)\sim\frac x{n^2}$$
and hence it converges for all $x>-1$ since $\sum1/n^2$ converges.
A: HINT:
Note that for $x\ge 0$, $n\ge 1$
$$\begin{align}
0\le \log\left(\sqrt[n]{1+\frac xn}\right)\le \frac{x}{n^2}
\end{align}$$
What happens for $-1<x< 0$?  Can you finish now?
A: You are almost there with the equivalent, but you never want to write $\sim 0$, which is imprecise at best, wrong at worst. The equivalent is supposed to give you how fast the convergence is. Moreover, the $\ln$ in the expression does not grow, so the sequence $(a_n)$ tends to zero because it is of the type $0/\infty$.
Now, getting back to this, you have $\ln(1+u) \sim u$ as $u \to 0$, and therefore
$$
a_n = \ln \sqrt[n]{1 + \frac{x}{n}} = \frac1n \ln \left ( 1 + \frac{x}{n} \right ) \sim \frac1n \frac{x}{n} = \frac{x}{n^2},
$$
so $|a_n| \sim |x|/n^2$. We know that the series $\sum |x|/n^2$ converges, so $\sum a_n$ converges absolutely.
It is true that technically, the expression is only valid for $n$ large enough, but you typically disregard this when you deal with series.
A: $$S=\lim_{n \infty} \sum_{k=1}^{n} \ln \sqrt[n]{1+\frac{k}{n}}=\lim_{n \infty} \frac{1}{n}\sum_{k=1}^{n} \ln (1+\frac{k}{n})= \int_{0}^{1}\ln(1+\frac{k}{n}) dx$$
$$S=(1+x)\ln(1+x)-(1+x)|_{0}^{1}=\ln 2-2+1= 2\ln 2-1$$
