Let $H_n=\sum _{k=1}^n \frac{1}{k}$ be the n-th harmonic number. Since $H_{k+1}>H_k$ for $k=1, 2, 3, ...$ the sequence $\frac{1}{H_k}$ is monotonic decreasing as $n \to \infty$, and the Leibniz criterion tells us that the alternating series
$$s_{H}=\sum _{k=1}^{\infty } \frac{(-1)^{k+1}}{H_k}\tag{1}$$
converges.
Its numerical value is
$$N(s_{H}) = 0.626332...$$
The natural question arises:
Question 1
Is there a closed form for $s_{H}$, i.e. an expression in terms of known constants (or is it even a new constant)? A possible selection of these constants might be those arising in the power series expansion of the zeta function, i.e.
$$const=\left\{\gamma ,\log (\pi ),\log (2),\gamma _1,\gamma _2,\zeta (3), ...\right\}$$
Question 2
A variation of the question replaces the harmonic number by the logarithm, starting at $k=2$, and asks for a closed form of the series
$$s_{L}=\sum _{k=2}^{\infty } \frac{(-1)^k}{\log (k)}\tag{2}$$
here
$$N(s_{L})=0.924299 ...$$
Solution attempts
For question 1 I have no idea to find a closed form, but the numerical value can be found to high precision.
Mathematica gives the first 100 Digits with this command:
NSum[(-1)^(n + 1)/HarmonicNumber[n], {n, 1, Infinity}, WorkingPrecision -> 100, Method -> "AlternatingSigns"]
$N(s_{H})=0.626332482737912354708657266227986063950088333562581965723069813694423327263764315345087698850095778034583404083688989609231701677593263$
I cannot tell if all these digits are correct.
My attempt to solve question 2 starts with the replacement
$$\frac{1}{\log (k)}=\int_0^{\infty } \exp (-t \log (k)) \, dt=\int_0^{\infty } k^{-t} \, dt$$
The summation under the integral is just the definition of the alternating zeta function starting at k = 2 which can be written as
$$\sum _{k=1}^{\infty } (-1)^k k^{-t}= (1-\zeta (t))+2^{1-t} \zeta (t)$$
Where $\zeta (t)=\sum _{k=1}^{\infty } k^{-t}$ is the Riemann zeta function.
Hence we find
$$s_{L}=\int_0^{\infty } ( (1-\zeta (t))+2^{1-t} \zeta (t) )\, dt\tag{3}$$
The integrand is well behaved in the region of integration (it resembles the decaying exponential).
But still I am stuck here (and Mathematica as well).
Table lookups
Stimulated by a comment of "J. M. is not a mathematician" I looked up the constants $s_{H}$ and $s_{L}$ defined here in the available tables.
The results are:
The Inverse Symbolic Calculator [1] could not identify the two constants.
The Online-Encyclopedia of integer sequences, searched for the sequence of the decimal digits, does not find $s_{H}$ but it does find $s_{L}$, and what's more, it contains the series of question 2: A099769 Decimal expansion of Sum_{n >= 2} (-1)^n/log(n). [2]
Here I.V.Blagouchine gives the following interesting integral representation without proof
$$s_{LB}=\int_0^{\infty } \frac{8 \tan ^{-1}(x)}{\sinh (2 \pi x) \left(\log ^2\left(4 x^2+4\right)+4 \tan ^{-1}(x)^2\right)} \, dx+\frac{1}{2 \log (2)}\tag{4}$$
Mathematica gives the first 100 digits with this command:
NSum[(-1)^n/Log[n], {n, 2, Infinity}, WorkingPrecision -> 100, Method -> "AlternatingSigns"]
$N(s_{L})=0.9242998972229388559595701813595900537733193978869190747796304372507005417114 3468979899134744193228$
Ths coincides with the digits given in the comment of robjohn and those of Ref. [2].
References
[1] https://isc.carma.newcastle.edu.au/index
[2] https://oeis.org/A099769
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 $$ \sum_{k=2}^\infty\frac{(-1)^k}{\log(k)} =0.92429989722293885595957018135959005377331939789 $$