Convergence of $\sum _{n=2}^{\infty }\:\frac{1}{n\ln\left(n\right)}$ Is this convergent or not?
$$\sum _{n=2}^{\infty }\:\frac{1}{n\ln\left(n\right)}$$
I tried using the ratio test but the limit is giving me 1, which doesn't help me. I don't think I'm supposed to use the integral test, since we haven't studied it.
 A: If you are familiar with Cauchy Condensation Test, the condens series becomes 
$$\sum_{n} 2^n \frac{1}{2^n \ln (2^n)}$$
If not, you can use the idea of the test:
$$\frac{1}{2 \ln2 } \geq \frac{1}{2 \ln 2 }  \\
\frac{1}{3 \ln 3 } \geq \frac{1}{4 \ln 2^2 }  \\
\frac{1}{4 \ln 4 } \geq \frac{1}{4 \ln 2^2 }  \\
\frac{1}{5 \ln 5 } \geq \frac{1}{8 \ln 2^3 }  \\
.....\\
\frac{1}{2^n \ln 2^n } \geq \frac{1}{2^n \ln 2^n } $$
By adding you get
$$s_{2^n} \geq \frac{1}{2 \ln 2 } +\frac{1}{2 \ln 2^2 } +...+\frac{1}{2 \ln 2^n }=\frac{1}{2 \ln 2} (1+\frac12+\frac13+...+\frac1n)$$ 
A: Define $x_n$ as :
$$x_n=\frac1{n\log n}\implies 2^nx_{2^n}=\frac{2^n}{2^n\log2^n}=\frac1{\log 2}\frac1n$$
You should be familiar with the condensation test.
A: People are mentioning the Cauchy Condensation Test, without saying what it is. Assuming you're not familiar with it, as a lot of people are not:
If $x_n>0$ and $x_{n+1}\le x_n$ then $\sum x_n<\infty$ if and only if $\sum_k2^kx_{2^k}<\infty$. Proving this is a good exercise; as has been pointed out, it shows your series diverges.
