Is there a closed form for the alternating series of inverse harmonic numbers?

I don't compute a closed form, but at the request of Dr. Wolfgang Hintze, I am supplying the computations involved in computing $47$ decimal places of $$ \sum_{k=2}^\infty\frac{(-1)^k}{\log(k)} $$ The Euler-Maclaurin Sum Formula says that $$\newcommand{\li}{\operatorname{li}} \begin{align} \sum_{k=2}^n\frac1{\log(k)} &=C_1+\li(n)+\frac1{2\log(n)}-\frac1{12n\log(n)^2}\\ &+\frac1{n^3}\left(\frac1{360\log(n)^2}+\frac1{120\log(n)^3}+\frac1{120\log(n)^4}\right)\\ &-\frac1{n^5}\left(\scriptsize\frac1{1260\log(n)^2}+\frac5{1512\log(n)^3}+\frac1{144\log(n)^4}+\frac1{126\log(n)^5}+\frac1{252\log(n)^6}\right)\\ &+\frac1{n^7}\left(\tiny\frac1{1680\log(n)^2}+\frac7{2400\log(n)^3}+\frac{29}{3600\log(n)^4}+\frac7{480\log(n)^5}+\frac5{288\log(n)^6}+\frac1{80\log(n)^7}+\frac1{240\log(n)^8}\right)\\ &-\frac1{n^9}\left(\tiny\frac1{1188\log(n)^2}+\frac{761}{166320\log(n)^3}+\frac{29531}{1995840\log(n)^4}+\frac{89}{2640\log(n)^5}+\frac{1069}{19008\log(n)^6}+\frac3{44\log(n)^7}\right.\\ &\phantom{-\frac1{n^9}\left(\vphantom{\frac1{\log(n)}}\right.}\left.\tiny+\frac{91}{1584\log(n)^8}+\frac1{33\log(n)^9}+\frac1{132\log(n)^{10}}\right)\\ &+O\left(\frac1{n^{11}\log(n)^2}\right) \end{align} $$ and $$ \begin{align} \sum_{k=1}^n\frac1{\log(2k)} &=C_2+\frac12\li(2n)+\frac1{2\log(2n)}-\frac1{12n\log(2n)^2}\\ &+\frac1{n^3}\left(\frac1{360\log(2n)^2}+\frac1{120\log(2n)^3}+\frac1{120\log(2n)^4}\right)\\ &-\frac1{n^5}\left(\scriptsize\frac1{1260\log(2n)^2}+\frac5{1512\log(2n)^3}+\frac1{144\log(2n)^4}+\frac1{126\log(2n)^5}+\frac1{252\log(2n)^6}\right)\\ &+\frac1{n^7}\left(\tiny\frac1{1680\log(2n)^2}+\frac7{2400\log(2n)^3}+\frac{29}{3600\log(2n)^4}+\frac7{480\log(2n)^5}+\frac5{288\log(2n)^6}+\frac1{80\log(2n)^7}+\frac1{240\log(2n)^8}\right)\\ &-\frac1{n^9}\left(\tiny\frac1{1188\log(2n)^2}+\frac{761}{166320\log(2n)^3}+\frac{29531}{1995840\log(2n)^4}+\frac{89}{2640\log(2n)^5}+\frac{1069}{19008\log(2n)^6}+\frac3{44\log(2n)^7}\right.\\ &\phantom{-\frac1{n^9}\left(\vphantom{\frac1{\log(n)}}\right.}\left.\tiny+\frac{91}{1584\log(2n)^8}+\frac1{33\log(2n)^9}+\frac1{132\log(2n)^{10}}\right)\\ &+O\left(\frac1{n^{11}\log(n)^2}\right) \end{align} $$ where the log-integral is defined as $$ \li(x)=\int_0^x\frac{\mathrm{d}t}{\log(t)} $$ In any case, the log-integral disappears in the alternating sum $$ \begin{align} \sum_{k=2}^{2n}\frac{(-1)^k}{\log(k)} &=2\sum_{k=1}^n\frac1{\log(2k)}-\sum_{k=2}^{2n}\frac1{\log(k)}\\ &=C_3+\frac1{2\log(2n)}-\frac1{8\,n\log(2n)^2}\\ &+\frac{15}{8\,n^3}\left(\frac1{360\log(2n)^2}+\frac1{120\log(2n)^3}+\frac1{120\log(2n)^4}\right)\\ &-\frac{63}{32\,n^5}\left(\scriptsize\frac1{1260\log(2n)^2}+\frac5{1512\log(2n)^3}+\frac1{144\log(2n)^4}+\frac1{126\log(2n)^5}+\frac1{252\log(2n)^6}\right)\\ &+\frac{255}{128\,n^7}\left(\tiny\frac1{1680\log(2n)^2}+\frac7{2400\log(2n)^3}+\frac{29}{3600\log(2n)^4}+\frac7{480\log(2n)^5}+\frac5{288\log(2n)^6}+\frac1{80\log(2n)^7}+\frac1{240\log(2n)^8}\right)\\ &-\frac{1023}{512\,n^9}\left(\tiny\frac1{1188\log(2n)^2}+\frac{761}{166320\log(2n)^3}+\frac{29531}{1995840\log(2n)^4}+\frac{89}{2640\log(2n)^5}+\frac{1069}{19008\log(2n)^6}+\frac3{44\log(2n)^7}\right.\\ &\phantom{-\frac1{512n^9}\left(\vphantom{\frac1{\log(n)}}\right.}\left.\tiny+\frac{91}{1584\log(2n)^8}+\frac1{33\log(2n)^9}+\frac1{132\log(2n)^{10}}\right)\\ &+O\left(\frac1{n^{11}\log(n)^2}\right) \end{align} $$ Plug in $n=10000$ and we get $C_3$ to over $45$ places $$ \sum_{k=2}^\infty\frac{(-1)^k}{\log(k)} =0.92429989722293885595957018135959005377331939789 $$