Show that the limit $$ \lim_{N\rightarrow\infty}\bigg(\sum_{k=1}^{2N-1}(-1)^{k+1}\frac{\sin\big(\frac{\pi}{2N}\big)}{\sin\big(\frac{k\pi}{2N}\big)}\bigg) = 2\ln2 $$ holds.
Hint: I think the identity $$\frac{1}{\sin(z\pi)}=\frac1\pi\sum_{n=-\infty}^{\infty}\frac{(-1)^n}{n+z}$$ may be of some help.
This is a thinly disguised version of the sum $1-\frac 12+\frac 13 -\frac 14+\cdots=\log 2$.
Split the sum into three ranges, (1) $1\le k\le N-1$, (2) $N+1\le k\le 2N-1$, and (3) $k=N$. Throw the third piece away as it vanishes in the limit as $N\to\infty$. After making the transformation $k\mapsto 2N-k$, (2) becomes identical to (1). This leaves us with the limit of $$2\left( \frac{\sin \alpha_N}{\sin \alpha_N}-\frac {\sin \alpha_N}{\sin 2\alpha_N}+\frac {\sin \alpha_N}{\sin 3\alpha_N}-\cdots \pm \frac{\sin \alpha_N}{\sin (N-1)\alpha_N} \right), \qquad \alpha_N:=\frac{\pi}{2N}. $$ The terms in the sum occur in pairs, each positive term paired with a succeeding negative term. If there is an unpaired term at the end, throw it away as it will vanish in the limit as $N\to\infty$. Then, subtract the terms in each pair to get $$ \lim_{N\to\infty} 2\sum_{k \ge 1} b_{N,k}, \quad (*) $$ where $b_{N,k}$ is zero if $k>\lfloor (N-1)2 \rfloor$ and otherwise $$b_{N,k}:=\frac{\sin{\alpha_N} (\sin{(2k\alpha_N)} - \sin{((2k-1)\alpha_N)})}{\sin{((2k-1)\alpha_N)} \sin{(2k\alpha_N)}}. $$ Notice that for each fixed $k$, we have $$ \lim_{N\to\infty} b_{N,k} = \frac{1}{(2k-1)(2k)} $$ so \begin{eqnarray*} 2 \sum_{k\ge 1} \lim_{N\to\infty} b_{N,k} &=& 2\left(\frac{1}{1\cdot 2} + \frac{1}{3\cdot 4} + \frac{1}{5\cdot 6} +\cdots\right)\\ &=&2 \left(1-\frac 1 2 + \frac 13 -\frac 14+\cdots\right)=2\log 2. \end{eqnarray*} Therefore, if the limit and the sum in $(*)$ could be interchanged, you would have what you want. Use the dominated convergence theorem to argue that they can be interchanged.