Does limit exist for the following expression?

Question

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 : $\lim\limits_{n \to \infty } \sum\limits_{k=1}^n - (-1)^{k} \ln \left(\frac{1}{k}\right)$

Answer 1

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.

Answer 2

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.