Series involving a logarithm: ${\sum_{n=2}^\infty} {1\over \ln^2n}$ $${\sum_{n=2}^\infty} {1\over \ln^2n}$$
Can I substitute ${x}$=${1\over \ln n}$ and using the integral test, set it up to be 
$${\lim_{t\to \infty}} \int_2^tx^2 dx$$ and solve from there? and then plug ${1\over \ln n}$ back in when I solve the integral in order to figure out whether it is convergent or divergent?
 A: Notice that
$$\frac1{\ln^2 n}\ge \frac1{n\ln n}$$
and the last series is divergent using the integral test
$$\int_2^\alpha \frac{dx}{x\ln x}=\ln(\ln x)\Bigg|_2^\alpha\xrightarrow{\alpha\to\infty}\infty$$
A: Do you have a guess? Remember that series of type $\sum \frac{1}{n^p}$ always diverge if $p \le 1$. 
It is a good thing to have in mind relations like theses:
$\log^p(n) << n^q << e^n << n!, p,q >0$
The symbol $f(n)<<g(n)$ means that $\lim_{n \rightarrow \infty} \frac{f(n)}{g(n)}=0$.
Now, $\frac{1}{ln^2(n)} > \frac{1}{n}$ by the comparison test you have your series diverges, because $\sum_n \frac{1}{n}$ diverges.
If you have any doubt why $\frac{1}{ln^2(n)} > \frac{1}{n}$, determine the limit $\lim_{n \rightarrow \infty} \frac{\log^2(n)}{n}$ using L'Hôpital rule, for example.
A: As stated, your integral is set up incorrectly. To use the integral test you need to calculate $$\int_2^\infty \frac{1}{\ln^2(n)}dn$$ So if you substitute $x = \frac{1}{\ln(n)}$ then you should find $$dx = \frac{-1}{n\ln^2(n)}dn \\ \implies dn = -n\ln^2(n)dx$$ where you can use your substitution to find $n=e^{1/x}, \space \ln^2(n) = \frac{1}{x^2}$ so $$dn = -\frac{e^{1/x}}{x^2}dx$$ and lastly $$\int_2^\infty \frac{1}{\ln^2(n)}dn = \int_2^\infty x^2\
\left(\frac{-e^{1/x}}{x^2}dx\right) \\ = -\int_2^\infty e^{1/x}dx$$ This should be an equivalent integral, although to solve this one I feel like you may want to do another substitution yet. Or better yet, skip the integral test and use the comparison test as Sami Ben Romdhane has suggested.
