TITLE: Estimating the sum $\sum_{k=2}^{\infty} \frac{1}{k \ln^2(k)}$
QUESTION [9 upvotes]: 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!

REPLY [3 votes]: 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
$$
\begin{align}
\sum_{k=2}^\infty\frac1{k\log(k)^2}
&=C\\
&=2.109742801236891974479257197616551326\tag2
\end{align}
$$