If limit exists, then what is its value? And if it does not exist then can we find where does this expression tends as $ n \to \infty$.
The expression : $\lim\limits_{n \to \infty } \sum\limits_{k=1}^n - (-1)^{k} \ln \left(\frac{1}{k}\right)$
For a sum $\sum_{k=1}^\infty a_k$ to converge, a necessary condition is that
$$\lim_{k \to \infty} a_k = 0,$$
which is not the case here. Therefore, your sum diverges.
The sum can be simplified as
$\sum\limits_{k=1}^n - (-1)^{k} \ln \left(\frac{1}{k}\right) =\sum\limits_{k=1}^n (-1)^{k} \ln k $.
This sum diverges. However, let's see how it behaves for even and odd numbre of terms.
For an even number of terms,
$\begin{array}\\ S_{2n} &=\sum\limits_{k=1}^{2n} (-1)^{k} \ln k\\ &=\sum\limits_{k=1}^{n} (- \ln (2k-1)+\ln(2k))\\ &=\sum\limits_{k=1}^{n} (- \ln (\frac{2k-1}{2k})\\ &=-\sum\limits_{k=1}^{n} \ln (1-\frac{1}{2k})\\ &\approx-\sum\limits_{k=1}^{n} -(\frac{1}{2k}+\frac{1}{8k^2}+ ...)\\ &=\sum\limits_{k=1}^{n} \frac{1}{2k}+\sum\limits_{k=1}^{n}(\frac{1}{8k^2}+ ...)\\ &=\frac12 H_n+C \quad\text{for some constant } C \text{ and large } n\\ &\approx\frac12 (\ln(n)+\gamma)+C\\ &=\frac12 \ln(n)+C_1\\ \end{array} $
Note that $C_1 =C+\frac12 \gamma $ and $C = \sum_{k=1}^{\infty} \sum_{m=2}^{\infty}\frac1{m(2k)^m } = \sum_{m=2}^{\infty}\frac1{m2^m}\sum_{k=1}^{\infty} \frac1{ k^m } = \sum_{m=2}^{\infty}\frac1{m2^m}\zeta(m) $ which converges and is easily computed.
For an odd number of terms,
$\begin{array}\\ S_{2n+1} &= S_{2n}-\ln(2n+1)\\ &\approx \frac12 \ln(n)+C_1-\ln(2n+1)\\ &= \frac12 \ln(n)+C_1-(\ln 2 + \ln(n+\frac12))\\ &= \frac12 \ln(n)+C_1-(\ln 2 + \ln(n)+\ln(1+\frac1{2n}))\\ &= -\frac12 \ln(n)+C_1-\ln 2+\frac1{2n}+O(\frac1{n^2}))\\ &\approx -\frac12 \ln(n)+C_1-\ln 2\\ \end{array} $
If we average the even and odd sums, $\frac12(S_{2n}+S_{2n+1}) \approx C_1-\frac12\ln 2 $.
This can be considered as the limit of the sum and $C_1$ can be computed as described previously.