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 
 A: For the first series,
$$ \ln(n)^{\ln(n)}=e^{\ln(n)\ln(\ln(n))}=n^{\ln(\ln(n))}$$
which grows faster than $n^p$ for any $p$. Therefore the series converges. 
Your argument for the second series looks good to me.
A: There is another way forward for the second problem.  In fact, we proceed using only elementary inequalities.  
In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities 
$$\frac{x-1}{x}\le\log(x)\le x-1 \tag 1$$
By letting $x=1+1/n$ in $(1)$ we obtain 
$$0\le \frac1n - \log\left(1+\frac1n\right)\le \frac{1}{n(n+1)}<\frac1{n^2}$$
Since $\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}$, then by the comparison test, the series of interest converges.

For the first problem, we can use the integral test.  Let $f(n)=\frac{1}{\log(n)^{\log(n)}}$.  Then, 
$$\begin{align}
\int_1^\infty \frac{1}{\log(x)^{\log(x)}}\,dx&=\int_0^\infty \left(\frac{e}{x}\right)^x\,dx \tag 2 \end{align}$$
Since $\left(\frac{e}{x}\right)^x\le \left(\frac{e}{x}\right)^2$ for $x\ge 2$, the integral in $(2)$ converges and therefore, by the integral test, the series of interest in the first problem does likewise.
A: 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}
$
