Estimating the sum $\sum_{k=2}^{\infty} \frac{1}{k \ln^2(k)}$

\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.)

I know the question is to show that the sum is greater than $2$, and Greg Martin's answer does that perfectly; however, I thought it might be interesting to compute the actual value of the sum.

Using the Euler-Maclaurin Sum Formula, we get $$ \begin{align} &\sum_{k=2}^n\frac1{k\log(k)^2}\\ &=C-\frac1{\log(n)}+\frac1{2n\log(n)^2}-\frac1{12n^2}\left(\frac1{\log(n)^2}+\frac2{\log(n)^3}\right)\\ &+\frac1{360n^4}\left(\frac3{\log(n)^2}+\frac{11}{\log(n)^3}+\frac{18}{\log(n)^4}+\frac{12}{\log(n)^5}\right)\\ &\scriptsize\,-\,\frac1{15120n^6}\left(\frac{60}{\log(n)^2}+\frac{274}{\log(n)^3}+\frac{675}{\log(n)^4}+\frac{1020}{\log(n)^5}+\frac{900}{\log(n)^6}+\frac{360}{\log(n)^7}\right)\\ &\tiny\,+\,\,\frac1{50400n^8}\left(\frac{210}{\log(n)^2}+\frac{1089}{\log(n)^3}+\frac{3283}{\log(n)^4}+\frac{6769}{\log(n)^5}+\frac{9800}{\log(n)^6}+\frac{9660}{\log(n)^7}+\frac{5880}{\log(n)^8}+\frac{1680}{\log(n)^9}\right)\\ &\tiny\,-\,\,\frac1{1995840n^{10}}\left(\frac{15120}{\log(n)^2}+\frac{85548}{\log(n)^3}+\frac{293175}{\log(n)^4}+\frac{723680}{\log(n)^5}+\frac{1346625}{\log(n)^6}+\frac{1898190}{\log(n)^7}+\frac{1984500}{\log(n)^8}+\frac{1461600}{\log(n)^9}+\frac{680400}{\log(n)^{10}}+\frac{151200}{\log(n)^{11}}\right)\\ &+O\!\left(\frac1{n^{12}\log(n)^2}\right)\tag1 \end{align} $$ Therefore, plugging $n=1000$ into $(1)$, we get $$ \begin{align} \sum_{k=2}^\infty\frac1{k\log(k)^2} &=C\\ &=2.109742801236891974479257197616551326\tag2 \end{align} $$