How to determine if the following series converge or not? $\Sigma_{n=1}^{\infty} a_n $ where:


*

*$ a_n = \frac{1}{\ln(n)^{\ln(n)}}$

*$a_n = \frac{1}{n }-\ln\left( 1+\frac{1}{n}\right)$
in the first case, I really have no idea
in the second case, is it correct to say that for $ \frac{1}{n }-\ln\left( 1+\frac{1}{n}\right)$ is (by taylor expansion) $\frac{1}{2n^2}+O(\frac{1}{n^3})$ and therefore, by the limit comparison test  converges?Is there any other way?
Thanks in advance 

For
2.
$\begin{array}\\
a_n 
&= \frac{1}{n }-\ln\left( 1+\frac{1}{n}\right)\\
&= \frac{1}{n }-\int_1^{1+1/n} \frac{dx}{x}\\
&= \frac{1}{n }-\int_0^{1/n} \frac{dx}{1+x}\\
&= \int_0^{1/n} (1-\frac{1}{1+x})dx\\
&= \int_0^{1/n} (\frac{x}{1+x})dx\\
&< \int_0^{1/n} x\,dx\\
&= \frac{x^2}{2}|_0^{1/n}\\
&= \frac{1}{2n^2}\\
\end{array}
$
For a lower bound,
from the integral,
$a_n > 0$.
To be more precise,
$\begin{array}\\
a_n 
&= \int_0^{1/n} (\frac{x}{1+x})dx\\
&\gt \frac1{1+1/n}\int_0^{1/n} x\,dx\\
&= \frac1{1+1/n}\frac{x^2}{2}|_0^{1/n}\\
&= \frac1{1+1/n}\frac{1}{2n^2}\\
&= \frac{1}{2n(n+1)}\\
\end{array}
$
