How to prove $\sqrt[n+1]{n+1}-\sqrt[n]{n}\sim-\frac{\ln{n}}{n^2}$ 
Show that 
$\sqrt[n+1]{n+1}-\sqrt[n]{n}\sim-\frac{\ln{n}}{n^2}$,
when $n\to+\infty$

I'm learning Taylor's Formula. The given solution is:

$\sqrt[n+1]{n+1}-\sqrt[n]{n}=e^{\frac{\ln(n+1)}{n+1}}-e^{\frac{\ln{n}}{n}}$
and use Taylor's Formula:
$e^{\frac{\ln(n+1)}{n+1}}=1+\frac{\ln(n+1)}{n+1}+\frac{1}{2}(\frac{\ln(n+1)}{n+1})^2+\frac{1}{6}(\frac{\ln(n+1)}{n+1})^3+o(\frac{(\ln n)^3}{n^3})$ and 
$e^{\frac{\ln(n)}{n}}=1+\frac{\ln(n)}{n}+\frac{1}{2}(\frac{\ln(n)}{n})^2+\frac{1}{6}(\frac{\ln(n)}{n})^3+o(\frac{(\ln n)^3}{n^3})$,
then $\sqrt[n+1]{n+1}-\sqrt[n]{n}=-\frac{\ln{n}}{n^2}+o(\frac{\ln{n}}{n^2})\sim\frac{\ln n}{n^2}$

I don't understand the last step: why is it $=-\frac{\ln{n}}{n^2}+o(\frac{\ln{n}}{n^2})$ ? And any other solutions?
 A: You just need to handle the subtraction of series for $\sqrt[n+1]{n+1}$ and $\sqrt[n] {n} $ term by term in a proper manner upto 3 terms. The first term $1$ cancels out in both series. The second terms upon subtraction lead to
\begin{align}
A &= \frac{\log(n+1)}{n+1}-\frac{\log n} {n} \notag\\
&= \frac{\log(n+1)}{n+1}-\frac{\log n} {n+1} +\log n\left(\frac{1}{n+1}-\frac{1}{n}\right) \notag\\
&= \frac{1}{n+1}\log\left(1+\frac{1}{n}\right)-\frac{\log n}{n(n+1)}\notag\\
&= - \frac{\log n} {n^{2}}+\left(\frac{1}{n}-\frac{1}{n^{2}}+o\left(\frac{1}{n^{2}}\right)\right)\left(\frac{1}{n}-\frac{1}{2n^{2}}+o\left(\frac{1}{n^{2}}\right)\right)+\frac{\log n} {n} \left(\frac{1}{n}-\frac{1}{n+1}\right)\notag\\
&= - \frac{\log n} {n^{2}}+\frac{1}{n^{2}}+\frac{\log n} {n^{2}(n+1)}+o\left(\frac{1}{n^{2}}\right)\notag\\
&= - \frac{\log n} {n^{2}}+o\left(\frac{\log n} {n^{2}}\right)\notag 
\end{align}
The third terms on subtraction lead to 
\begin{align}
B &= \frac{A} {2}\left(\frac{\log(n+1)}{n+1}+\frac{\log n} {n} \right) \notag\\
&= o\left(\frac{\log n} {n^{2}}\right)\notag
\end{align} 
and therefore we have $$\sqrt[n+1]{n+1}-\sqrt[n]{n}=-\frac{\log n} {n^{2}}+o\left(\frac{\log n} {n^{2}}\right)$$
