Will this series converge, will its limit exist (and if so, what is it), and will its sum be $> 0$ or $< 0$? The series : $$\lim\limits_{n \to  \infty }  \sum\limits_{k=1}^n - (-1)^{k} \ln (k)$$
We know that, for a sum $\sum\limits_{k=1}^\infty a_k$ to converge, a necessary condition is that : 
$$\lim_{k\to\infty} a_k = 0$$
We can write the above series as 
$$\ln(1)-\ln(2)+\ln(3)-\ln(4)+\ln(5)-\ln(6)\cdots\infty$$
It becomes : \begin{align}
& =\ln\left(\frac{1}{2}\right) + \ln\left(\frac{3}{4}\right) + \ln\left(\frac{5}{6}\right) + \ln\left(\frac{7}{8}\right) + \ln\left(\frac{9}{10}\right) + \cdots\ln\left(\bf{\frac{n}{n+1}}\right)\\
\end{align}
and $$\text{as,}\quad n\to\infty\qquad\qquad\lim\limits_{n\to\infty}\left(\bf{\frac{n}{n+1}}\right) = 1$$ 
thus, $$\lim\limits_{n \to \infty}\bf{ln}\left(\bf{\frac{n}{n+1}}\right)=0\text{,}$$
that is, $$\lim\limits_{k\to\infty} a_k = 0\text{,}$$ 
So it should converge, we also observe that all the terms inside $log$ are less than 1 so all the individual $\textit{log }$ values would be negative and thus making the sum of series negative.
We also observe that as $n\to\infty$ the values inside $log$ will tend to $0$ from negative side(because each individual $log$ value is negative)thus, the absolute value of individual $log$ values would slowly decrease, with the greatest absolute value of $ln\left(\bf{\frac{1}{2}}\right)$
But we can write the above series in another way :
\begin{align}
& = \ln(1)-ln(2)+\ln(3)-\ln(4)+\ln(5)-\ln(6)\cdots\infty\\
& = \qquad-\ln(2)+\ln(3)-\ln(4)+\ln(5)-\ln(6)\cdots\infty\qquad\qquad\{\ln(1)=0\}
\end{align}
It becomes :
$$\ln\left(\frac{3}{2}\right) + \ln\left(\frac{5}{4}\right) + \ln\left(\frac{7}{6}\right) + \ln\left(\frac{9}{8}\right) + \ln\left(\frac{11}{10}\right) + \cdots\ln\left(\bf{\frac{n+1}{n}}\right)$$
Again,
$$\text{as,}\quad n\to\infty\qquad\qquad\lim\limits_{n\to\infty}\left(\bf{\frac{n+1}{n}}\right) = 1$$ 
thus, $$\lim\limits_{n \to \infty}\bf{ln}\left(\bf{\frac{n+1}{n}}\right)=0\text{,}$$
that is, $$\lim\limits_{k\to\infty} a_k = 0\text{,}$$ 
So it should converge, we also observe that all the terms inside $log$ are greater than 1 so all the individual $\textit{log }$ values would be positive and thus making the sum of series positive.
Now this is where it CONTRADICTS our above statement that the sum of series would be negative
We also observe that as $n\to\infty$ the values inside $log$ will tend to $0$ from positive side(because each individual $log$ value is positive)thus, the absolute value of individual $log$ values would slowly decrease, with the greatest absolute value of $ln\left(\bf{\frac{3}{2}}\right)$
One may further argue that in our original expression:
$$\lim\limits_{n \to  \infty }  \sum\limits_{k=1}^n - (-1)^{k} \ln (k)$$
the series does not Converge as our necessary condition,
$$\lim\limits_{k\to\infty} a_k = 0$$ 
does not satisfy because $- (-1)^{n} \ln (n)$ tends to $\infty$ or $-\infty$ depending on the value of $'n'$ as $n\to\infty$, thus contradicting our another statement that : 'this series converges'
Now at last, how do we calculate the limit of this series (even an approximate value).
Edit : What would be the answer to the same question if the series would have been:
$$\lim\limits_{n \to  \infty }  \sum\limits_{k=1}^n - (-1)^{k} \ln \left(\frac{1}{k}\right)$$

First of all, $\sum_{k=1}^\infty -(-1)^k\ln k$ does not converge: The summand $-(-1)^k\ln k$ does not converge to $0$. Instead you considered the different series
$$ \sum_{k=1}^\infty\left(\ln(2k-1)-\ln(2k)\right)$$
which is related to the original series by the fact that every second partial sum of the original is a partial sum of this new series. As you noted, we have at least $\ln(2k-1)-\ln(2k)\to 0$. However, for large $k$ we have $\ln(2k-1)-\ln(2k)=\ln\left(1-\frac1{2k}\right)\approx -\frac1{2k}$, so that the series (still) diverges - per comparison with thge harmonic series.
