Find $\left \lfloor 1000S \right\rfloor$ where $S=\sum_{k=2}^\infty (-1)^k\log_k e$ 
$$S =  \log_{2}{e} - \log_{3}{e} + \log_{4}{e} - \log_{5}{e} + \log_{6}{e}\cdots  $$

Find the Value of $\left \lfloor 1000S \right\rfloor$.

My Attempt
I changed it into summation and tried to form a Taylor series of $ \ln{(\ln{x} + 1)} $. But got nothing. Please help me in this.
 A: First, we can rewrite the sum as
$$\sum_{k=2}^\infty \frac{(-1)^k}{\log k} = \sum_{k=1}^\infty a_k
\quad\text{ where }\quad a_k \stackrel{def}{=} \left(\frac{1}{\log(2k)} - \frac{1}{\log(2k+1)}\right)$$
For each $k$, we can apply MVT to $\frac{1}{\log x}$ on the interval $[2k,2k+1]$ and
find a $\xi_k \in (0,1)$ such that
$$a_k = \phi(2k+\xi_k) \quad\text{ where }\quad
\phi(x) = \left(-\frac{1}{\log x}\right)' = \frac{1}{x(\log x)^2}$$ 
Notice $\phi(x)$ is monotonic decreasing for $x > 1$. For $k > 1$, we have
$$\int_k^{k+1}\phi(2x-2) dx \ge a_k = \phi(2k+\xi_k) \ge \int_k^{k+1}\phi(2x+1) dx$$
This leads to
$$\begin{align}
& \int_N^\infty \phi(2x-2) dx \ge \sum_{k=N}^\infty a_k \ge \int_N^{\infty} \phi(2x+1) dx\\
\implies &
\frac{1}{2\log(2N-2)} \ge \sum_{k=N}^{\infty} a_k \ge \frac{1}{2\log(2N+1)}
\end{align}
$$
This leads to following approximation of the sum $S$
$$S = \sum_{k=1}^\infty a_k \approx \sum_{k=1}^{N-1} a_k + \frac{1}{2\log(2N)}\tag{*1}$$
and the error is smaller than
$$\frac{1}{2\log(2N-2)} - \frac{1}{2\log(2N+1)} \le \frac{3}{2}\phi(2N-2)\tag{*2}$$
For $N = 150$, 


*

*RHS of $(*1)$ evaluates to $0.9242742823890622...$

*RHS of $(*2)$ evaluates to $0.0001550844673734...$


This is enough to conclude the first three digits of $S$ is $0.924$ and hence
$\lfloor 1000S \rfloor = 924$.
BTW, it seems $S$ has a closed form expression
$$\frac{2}{\log 2} + \int_0^\infty (2^{1-t}-1)(\zeta(t)-1) dt\tag{*3}$$
where $\zeta(t)$ is the Riemann Zeta function. However, I'm not sure whether this has any use
in computing $S$ accurately.
Update
About how to arrive at the expression $(*3)$, 
the hand waving argument is
$$\begin{align}
\sum_{k=2}^\infty \frac{(-1)^k}{\log k}
= & \sum_{k=2}^\infty (-1)^k \int_0^\infty k^{-t} dt\\
= & \int_0^\infty \left( 2^{-t} - 3^{-t} + 4^{-t} - \cdots \right) dt\\
= & \int_0^\infty \left( 2(2^{-t} + 4^{-t} + 6^{-t} + \cdots) - (2^{-t} + 3^{-t} + \cdots )\right) dt \\
= & \int_0^\infty \left(2^{1-t}\zeta(t) - (\zeta(t) - 1)\right) dt\\
= &\frac{2}{\log 2} + \int_0^\infty (2^{1-t} - 1)(\zeta(t) - 1) dt
\end{align}
$$
To make this rigorous, we zeta function regularize the original series.
For $\Re s >1$, a derivation likes above allow us to establish following 
identity rigorously.
$$\sum_{k=2}^\infty \frac{(-1)^k}{\log k} k^{-s} = \int_s^{\infty} \left[2^{1-t}\zeta(t) - (\zeta(t) - 1)\right] dt$$
We then argue we can analytic continue the last expression to $s = 0$ and 
justify $(*3)$.
