Another transposed harmonic alternating series

Let us use the notation $H_n=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}.$ The number $H_n$ is called the $n^{th}$ - harmonic number. If we add the first $(c+1)n$ first terms of the given series, we obtain the following expression $S_{(c+1)n}=\sum_{k=1}^{cn} \frac{1}{2k}-\sum_{k=1}^{n}\frac{1}{2k-1}=\sum_{k=1}^{cn} \frac{1}{2k}-\sum_{k=1}^{n}\frac{1}{2k}-\sum_{k=1}^{n}\frac{1}{2k-1}+\sum_{k=1}^{n}\frac{1}{2k}.$ $=\sum_{k=1}^{cn} \frac{1}{2k}-\sum_{k=1}^{2n}\frac{1}{k}+\sum_{k=1}^{n}\frac{1}{2k}.$ Thus we have got that: $S_{(c+1)n}=\frac{1}{2}H_{cn}-H_{2n}+\frac{1}{2}H_n.$ Therefore, $S_{(c+1)n}=\frac{1}{2}(H_{cn}-\ln{cn})-(H_{2n}-\ln{2n})+\frac{1}{2}(H_n-\ln n)+\frac{1}{2}\ln{cn}-\ln{2n}+\frac{1}{2}\ln{n}$ $=\frac{1}{2}(H_{cn}-\ln{cn})-(H_{2n}-\ln{2n})+\frac{1}{2}(H_n-\ln n)+\frac{1}{2}\ln{\frac{c}{4}}.$ Now using that $\lim_{n\rightarrow \infty}(H_n-\ln n)=\gamma,$ where $\gamma$ is the Euler–Mascheroni constant, and that the sum of the given series is $\frac{1}{2}\ln\frac{5}{4}$ , we obtain that $\lim_{n\rightarrow \infty} S_{(c+1)n}=\frac{1}{2}\ln{\frac{c}{4}}=\frac{1}{2}\ln\frac{5}{4},$ and therefore is easy to see that $\boxed{c=5}.$ For our solution to be complete, we would need to prove that our series converges when $c=5.$ The previous calculations show that $\lim_{n\rightarrow \infty}S_{6n}=\frac{1}{2}\ln\frac{5}{4}.$ To prove that $S_n$ converges to the same value, it would be enough to show that any subsequence of the form $S_{6n+r},$ where $r\in\{1, 2, 3, 4, 5\},$ converges also to $\frac{1}{2}\ln\frac{5}{4}$ , which is not difficult to obtain if you realize that $\lim_{n\rightarrow \infty}(S_{6n+r}-S_{6n})=0$ for any such a number $r.$

Compañero Presa has written a fine solution. For the sake of variety, let me group the terms in a different way:

Let $S_n$ be the partial sum up to the term $-\frac{1}{2n-1}$ , for a fixed $c$ . Then $S_n=\sum_{k=1}^{2n}\frac{(-1)^k}{k}+\sum_{k=n+1}^{cn}\frac{1}{2k}$ . The first summand tends to $-\ln(2)$ . The second summand is a right Riemann sum of $\frac{1}{2x}$ for $1\leq x\leq c$ , tending to $\frac{\ln(c)}{2}$ . Thus $\lim_{n\to\infty}S_n=\frac{1}{2}\ln(\frac{c}{4})$ , and $c=\boxed{5}$ . (Since the terms of the series tend to 0, all the partial sum will have the same limit.)