Evaluating $\sum_{k=1}^{\infty} 2\ln{(2k)} - \ln{(2k-1)} - \ln{(2k+1)} $

Question

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?

Answer 1

Let $S_N$ be the sum of the first $N$ terms.

Then $$\exp(S_N)=\prod_{k=1}^N\frac{(2k)^2}{(2k-1)(2k+1)} =\frac{4^N N!^2}{(2N+1)(1\times 3\times 5\cdots\times (2N-1))^2} =\frac{16^N N!^4}{(2N+1)(2N)!^2}.$$ You can now attack this with Stirling's formula.

Answer 2

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.