By integral test, it is easy to see that $$\sum_{k=2}^{\infty} \frac{1}{k \ln^2(k)}$$ converges. [Here $\ln(x)$ denotes the natural logarithm, and $\ln^2(x)$ stands for $(\ln(x))^2$]
I am interested in proving the following inequality (preferrably using integral calculus) $$\sum_{k=2}^{\infty} \frac{1}{k \ln^2(k)}>2$$
By wolfram alpha, the actual value of the sum is about 2.10974. Since $\frac{1}{k \ln^2(k)}$ is decreasing, we have
$$
\sum_{k=2}^{\infty} \frac{1}{k \ln^2(k)}\ge \int_2^{\infty} \frac{1}{x \ln^2(x)}
dx=\frac{1}{\ln(2)}\approx 1.4427$$
So this lower bounded is weaker than desired.
My motivation for asking this question is that by being able to estimate this particular sum will hopefully teach me a general technique which I may try applying to sums of the form $\sum_{k=1}^{\infty} f(k)$.
Thanks!
\begin{align*} \sum_{k=2}^{\infty} \frac1{k\ln^2k} &= \frac1{2\ln^22} + \frac1{3\ln^23} + \sum_{k=4}^{\infty} \frac1{k\ln^2k} \\ &\ge \frac1{2\ln^22} + \frac1{3\ln^23} + \int_4^{\infty} \frac{dx}{x\ln^2x} \\ &= \frac1{2\ln^22} + \frac1{3\ln^23} + \frac{1}{\ln4} > 2.038. \end{align*} (I see Andrew just wrote this in a comment.)