<s>[INST] Evaluating $\sum_{k=1}^{\infty} 2\ln{(2k)} - \ln{(2k-1)} - \ln{(2k+1)} $ I was trying to evaluate the following series, which I know converges:
$$\sum_{k=1}^{\infty} 2\ln{(2k)} - \ln{(2k-1)} - \ln{(2k+1)} \tag{1}\label{1} $$
In a telescoping fashion, I began writing out the terms in hopes to find a pattern:
$$= (2\ln{2} - \ln{1} - \ln{3}) + (2\ln{4} - \ln{3} - \ln{5}) + (2\ln{6} - \ln{5} - \ln{7}) + \ldots \tag{2}\label{2}$$
while nothing canceled out, I grouped terms together:
$$ = 2\ln{2} - 2\ln{3} + 2\ln{4} - 2\ln{5} + 2\ln{6} - 2\ln{7} + \ldots \tag{3}\label{3}$$
$$ = 2 \left[ \ln{2} - \ln{3} + \ln{4} - \ln{5} + \ln{6} - \ln{7} + \ldots \right] \tag{4}\label{4}$$
which left me with the following divergent series:
$$ = 2 \sum _{k=2} ^{\infty} (-1)^k \ln{k} \tag{5}\label{5}$$
Clearly, $\eqref{5}$ can't be equivalent to $\eqref{1}$. 
I'm pretty new to calculus, and while I've covered telescoping series, it seems that this technique cannot be applied here. Though, I don't know why.
Where did I go wrong? 
 [/INST] Here is the answer of the series. With expansion of $\ln$ we have
\begin{align}
\sum_{k=1}^{\infty} 2\ln(2k)-\ln(2k-1)-\ln(2k+1) 
&= \sum_{k=1}^{\infty} -\ln\left(1-\dfrac{1}{2k}\right)-\ln\left(1-\dfrac{1}{2k}\right) \\
&= \sum_{k=1}^{\infty} \sum_{n\geqslant1} \dfrac{1}{n}\left(\dfrac{1}{2k}\right)^{2n} \\
&= \sum_{n\geqslant1} \dfrac{1}{n} \sum_{k=1}^{\infty} \left(\dfrac{1}{2k}\right)^{2n} \\
&= \sum_{n\geqslant1} \dfrac{\zeta(2n)}{n2^{2n}} \\
&= \color{blue}{\ln\dfrac{\pi}{2}}
\end{align}
the last step proved here.
</s>