$ \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.
 A: Take $f(x)=\ln\left(1+\frac{1}{4x^2}\right)$ over $[1,+\infty)$ and see that $f'(x)=-\frac{1}{2x^3(1+1/4x^2)}$ and then is decreasing over $[1,+\infty)$. $f(x)$ is also positive and continuous so you can use the integral test for $\displaystyle{\sum_{k=1}^{+\infty}\ln\left(1+\frac{1}{4k^2}\right)}$: 
$$\int_1^{+\infty}\ln\left(1+\frac{1}{4x^2}\right)dx$$ 
Using integral by parts, you have the following answer as an upper bound: 
$$\int_1^{+\infty}\ln\left(1+\frac{1}{4x^2}\right)dx=[-2\ln(2)x+x\ln(4+1/x^2)+\arctan(2x)]\big|_{x=1}^{+\infty}\\=2\ln(2)-\arctan(2)-\ln(5)+(1/2)\pi$$ 
For solving the integral take $\displaystyle{u=\ln\left(1+\frac{1}{4x^2}\right)}$ and $dv=dx$. :-)
A: The sum equals
$$
\ln\Bigl(\frac{2}{\pi}\sinh\Bigl(\frac{\pi}{2}\Bigr)\Bigr)=\ln\Bigl(\frac{e^{\pi/2}-e^{-\pi/2}}{\pi}\Bigr).
$$
It can be obtained evaluating at $z=i\pi/2$ the product formula for the sine function
$$
\sin z=z\,\prod_{k=1}^\infty\Bigl(1-\frac{z^2}{\pi^2n^2}\Bigr),\qquad z\in\mathbb{C}.
$$
I doubt that there is an elementary way to prove it.
A: The high school students I know would solve it like this.
A: 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
$$\begin{eqnarray}
   \sum_{k=1}^\infty \log\left(1+\frac{1}{4k^2}\right) &=& \sum_{n=0}^\infty \frac{(-1)^n}{n+1} \frac{\zeta(2n+2)}{4^{n+1}} = \sum_{n=1}^\infty \frac{\pi^{2n}}{2n} \frac{B_{2n}}{(2n)!} \\ &=& \sum_{n=1}^\infty \frac{\pi^{n}}{n} \frac{B_{n}}{(n)!} + \frac{\pi}{2} \\&=& \frac{\pi}{2} + \int_0^1 \left(\sum_{n=1}^\infty \pi^{n} t^{n-1}\frac{B_{n}}{(n)!}\right) \mathrm{d} t \\ &=& \frac{\pi}{2} + \int_0^1 \left(\frac{\pi}{\mathrm{e}^{\pi t}-1} - \frac{1}{t} \right) \mathrm{d} t \\ &=& \frac{\pi}{2} + \int_0^\pi \left(\frac{1}{\mathrm{e}^{u}-1} - \frac{1}{u} \right) \mathrm{d} u \\ &=&
  \frac{\pi}{2} + \left. \log\left(\frac{1-\mathrm{e}^{-u}}{u}\right) \right|_{0}^{\pi} = \frac{\pi}{2} + \log\left(\frac{1-\exp(-\pi)}{\pi}\right) = \log\left(\frac{2}{\pi} \cdot \sinh\frac{\pi}{2}\right)
\end{eqnarray} \tag{1}
$$
A: A possible solution:
$\displaystyle \sum\limits_{k=1}^{+ \infty} \ln \left( 1+ \frac{1}{4k^2} \right)= \ln \left( \prod\limits_{k=1}^{+ \infty}  \left( 1+ \frac{1}{4k^2} \right) \right)$. Using Euler's formula, $ \displaystyle \prod\limits_{k=1}^{+ \infty}  \left( 1+ \frac{1}{4k^2} \right)= -i\frac{2}{\pi} \sin \left( i \frac{\pi}{2} \right)= \frac{2}{\pi} \sinh \left( \frac{\pi}{2} \right) $.
Finally, $\displaystyle \sum\limits_{k=1}^{+ \infty} \ln \left(1+ \frac{1}{4k^2} \right)= \ln \left( \frac{2}{\pi} \sinh \left( \frac{\pi}{2} \right) \right)$.
