$\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 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.
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.
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} $
