Using the integral test for $\sum _{n=2}^\infty \frac{1}{(\ln n)^{\ln n}}$ I'm already aware that this series does, in fact, converge. But how can you prove it using the integral test? I was trying to compute$$\int \frac{1}{x^{\ln(\ln(x))}}\,dx,$$since
$(\ln n)^{\ln n}=n^{\ln \ln n}$, but didn't get a primitive. In fact, I was researching this problem and it's on Spivak's Calculus book. It's on the chapter of infinite series and as a hint, it says that one should prove that $\displaystyle \int \frac{e^y}{y^y}dy$ exists using $\displaystyle \sum \limits _{n=1}^\infty \left (\frac{e}{n}\right )^n$. Any idea or hint on how to use that?
 A: By the substitution $u=\ln x$, you can find
$$
\int_2^\infty \frac{1}{\ln x^{\ln x}}\mathrm dx=\int_{\ln 2}^\infty \frac{1}{u^u}e^u\mathrm du
$$
now eventually $u\geq e^2$, and you can estimate the tail pretty easily.
A: Take a look at
$$
\int^\infty_2 \frac{1}{\ln(x)^{\ln(x)}} ~\mathrm{d}x.
$$
Substitute $y = \ln(x)$ to get
$$
\int^\infty_{\ln(2)} \frac{\exp(y)}{y^{y}} ~\mathrm{d}y.
$$
Using the series test, this converges iff
$$
\sum_{n = 1}^\infty \frac{\exp(n)}{n^n} = \sum_{n = 1}^\infty \left(\frac{e}{n}\right)^n
$$
converges. We estimate
$$
\left \lvert \sum_{n = 1}^\infty \left(\frac{e}{n}\right)^n \right \rvert \leq \sum_{n = 1}^\infty \left(\frac{e}{n}\right)^n = \sum_{n = 1}^2\left(\frac{e}{n}\right)^n + \sum_{n = 3}^\infty\left(\frac{e}{n}\right)^n \leq \sum_{n = 1}^2\left(\frac{e}{n}\right)^n + \sum_{n = 3}^\infty\left(\frac{e}{3}\right)^n
$$
Since $0< \frac{e}{3} < 1$, the last series is a converging geometric series. So the series converges and so does your original series.
A: $$ \int_1^{\infty} \left( \frac{e}{y} \right)^y dy = \sum_{k=1}^{\infty} \underbrace{\left( \int_k^{k+1} \left( \frac{e}{y} \right)^y dy \right)}_{ \large{ \overset{*}{\leq} \left( \frac{e}{k} \right)^k } }$$
$$ \leq \sum_{k=1}^{\infty} \left( \frac{e}{k} \right)^k $$
$$ \leq \left( \frac{e}{1} \right)^1 + \left( \frac{e}{2} \right)^2 + \sum_{k=3}^{\infty} \left( \frac{e}{3} \right)^k,\ \text{ which converges.} $$
*because $f(x) = \left( \frac{e}{x} \right)^x$ is decreasing on $[1,\infty), $ which can be checked by looking at $f'(x).$
