$ \sum_{k=1}^{\infty} \ln{\left(1 + \frac{1}{4 k^2}\right)}$ Computing this sum Compute the limit:
$$ \sum_{k=1}^{\infty} \ln{\left(1 + \frac{1}{4 k^2}\right)}$$
My teacher says it can be solved by only using high school knowledge, but I don't see how. What did I try? Well, I thought of Riemann sums but I see no way to connect this sum to it. Thanks.
I'm only interested in a solution at high school level if possible !
UPDATE: Now, I'm my own teacher.

As an alternative to @Seiros solution, albeit far less elegant is to use $$\log\left(1+\frac{1}{4k^2}\right) = \sum_{n=0}^\infty \frac{(-1)^n}{n+1} \frac{1}{4^{n+1} k^{2n+2}}$$
Then
$$\begin{eqnarray}
   \sum_{k=1}^\infty \log\left(1+\frac{1}{4k^2}\right) &=& \sum_{n=0}^\infty \frac{(-1)^n}{n+1} \frac{\zeta(2n+2)}{4^{n+1}} = \sum_{n=1}^\infty \frac{\pi^{2n}}{2n} \frac{B_{2n}}{(2n)!} \\ &=& \sum_{n=1}^\infty \frac{\pi^{n}}{n} \frac{B_{n}}{(n)!} + \frac{\pi}{2} \\&=& \frac{\pi}{2} + \int_0^1 \left(\sum_{n=1}^\infty \pi^{n} t^{n-1}\frac{B_{n}}{(n)!}\right) \mathrm{d} t \\ &=& \frac{\pi}{2} + \int_0^1 \left(\frac{\pi}{\mathrm{e}^{\pi t}-1} - \frac{1}{t} \right) \mathrm{d} t \\ &=& \frac{\pi}{2} + \int_0^\pi \left(\frac{1}{\mathrm{e}^{u}-1} - \frac{1}{u} \right) \mathrm{d} u \\ &=&
  \frac{\pi}{2} + \left. \log\left(\frac{1-\mathrm{e}^{-u}}{u}\right) \right|_{0}^{\pi} = \frac{\pi}{2} + \log\left(\frac{1-\exp(-\pi)}{\pi}\right) = \log\left(\frac{2}{\pi} \cdot \sinh\frac{\pi}{2}\right)
\end{eqnarray} \tag{1}
$$
