Evaluate $\sum_{n=2}^{\infty} \frac{1}{n\log_2^2{n}}$ I would like to evaluate
$$\sum_{n=2}^{\infty} \frac{1}{n\log_{2}^{2}\left(n\right)}.
$$ I know that $\int_2^{\infty} \frac{1}{n\log_2^2{n}} = \log{2}$ so my guess is that the answer is close to that. However numerically it seems to be very close to $1$.    Is it in fact $1$?
Using maple limit(sum(1/(n*log[2](n)^2),n=2..i), i=infinity); gives
$$\frac{48 \ln^2{2} - 48 O(1) \ln{2} + 13\ln{2} + 2}{48 \ln{2}}$$
however I have no idea how to interpret this.
 A: It's straightforward to show that
$${1\over N(\ln N)^2}+{1\over(N+1)(\ln (N+1))^2}+{1\over(N+2)(\ln(N+2))^2}+\cdots\gt{1\over\ln N}+{1\over2N(\ln N)^2}$$
One thus obtains
$$\sum_{n=2}^\infty{1\over n\log_2^2n}=(\ln2)^2\sum_{n=2}^\infty{1\over n(\ln n)^2}\gt (\ln2)^2\left({1\over2(\ln2)^2}+{1\over\ln3}+{1\over6(\ln3)^2} \right)\approx1.00367$$
The "straightforward" inequality comes from comparing the series, thought of as a step function, to the integral beneath the curve $f(x)=1/x(\ln x)^2$:  The difference between the two sides is the area between the curve and the chords that connect points with integer $x$ values.  (The total area above those chords is $1/2N(\ln N)^2$.)
A: It's possible to rewrite this sum as an integral involving the Riemann zeta function.  Note that
$$
\zeta(z)-1=\sum_{n=2}^{\infty}n^{-z}=\sum_{n=2}^{\infty}\exp(-z\log n)
$$
when $\text{Re}(z)>1$.  Taking a single antiderivative (and treating convergence casually),
$$
\int_{z}^{\infty}(\zeta(s)-1)ds=\sum_{n=2}^{\infty}\int_{z}^{\infty}\exp(-s\log n)ds=\sum_{n=2}^{\infty}\frac{\exp(-z\log n)}{\log n}.
$$
Taking another one,
$$
\int_{z}^{\infty}dy \int_{y}^{\infty}(\zeta(s)-1)ds=\sum_{n=2}^{\infty}\int_{z}^{\infty}\frac{\exp(-y\log n)}{\log n}dy=\sum_{n=2}^{\infty}\frac{\exp(-z\log n)}{\log^2 n},
$$
or
$$
\sum_{n=2}^{\infty}\frac{n^{-z}}{\log^2 n}=\int_{z}^{\infty}dy\int_{y}^{\infty}(\zeta(s)-1)ds=\int_{z}^{\infty}(\zeta(s)-1)(s-z)ds.
$$
Here we want to evaluate this at $z=1$, and multiply by the correct factor to get the logarithm in base-$2$:
$$
\sum_{n=2}^{\infty}\frac{1}{n\log_2^2 n}=\log^2 2\int_{1}^{\infty}(\zeta(s)-1)(s-1)ds.
$$
