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.
Best Answer
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, $\displaystyle \sum\limits_{k=1}^{+ \infty} \ln \left(1+ \frac{1}{4k^2} \right)= \ln \left( \frac{2}{\pi} \sinh \left( \frac{\pi}{2} \right) \right)$.