\rm\prod_(k=0)^(\infty)(100k(k+1)+5^2)/(100k(k+1)+3^2) \\​

Question

`\rm\prod_(k=0)^(\infty)(100k(k+1)+5^2)/(100k(k+1)+3^2) \\`​

Answer 1

First rewrite the multiplicand as

`\displaystyle (100k(k+1)+5^2)/(100k(k+1)+3^2) = ((10k-5)^2)/((10k-1)(10k-9)) \\\\ ~~~~~~~~ = (((10k)^2 - 5^2)^2 (10k+1) (10k + 9))/(((10k)^2 - 1^2) ((10k)^2 - 9^2) (10k+5)^2) \\\\ ~~~~~~~~ = \frac{\left(1 - \left(\frac1{2k}\right)^2\right)^2 (10k+1) (10k+9)}{\left(1 - \left(\frac1{10k}\right)^2\right) \left(1 - \left(\frac9{10k}\right)^2\right) (10k+5)^2} \\\\ ~~~~~~~~ = \frac{\left(1 - \left(\frac1{2k}\right)^2\right)^2}{\left(1 - \left(\frac1{10k}\right)^2\right) \left(1 - \left(\frac9{10k}\right)^2\right)} \left(1 - (4^2)/(5^2(2k+1)^2)\right)`

Now recall the infinite product representations of
`\sin(x)` and
`\cos(x)`,

`\displaystyle \sin(x) = x \prod_(k=1)^\infty \left(1 - (x^2)/(\pi^2n^2)\right)`

`\displaystyle \cos(x) = \prod_(k=1)^\infty \left(1 - (4x^2)/(\pi^2(2n-1)^2)\right)`

from which we get

`\displaystyle \prod_(k=1)^\infty \left(1 - \left(\frac1{2k}\right)^2\right)^2 = \left(\prod_(k=1)^\infty \left(1 - (\left(\frac\pi2\right)^2)/(\pi^2 k^2)\right)\right)^2 \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~ = \left((\sin\left(\frac\pi2\right))/(\frac\pi2)\right)^2 = \frac4{\pi^2}`

`\displaystyle \prod_(k=1)^\infty \left(1 - \left(\frac1{10k}\right)^2\right) = \prod_(k=1)^\infty \left(1 - \frac{\left(\frac\pi{10}\right)^2}{\pi^2 k^2}\right) \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~ = \frac{\sin\left(\frac\pi{10}\right)}{\frac\pi{10}} \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~ = \frac{10}\pi \sin\left(\frac\pi{10}\right)`

`\displaystyle \prod_(k=1)^\infty \left(1 - \left(\frac9{10k}\right)^2\right) = \prod_(k=1)^\infty \left(1 - (\left((9\pi)/(10)\right)^2)/(\pi^2 k^2)\right) \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~ = (\sin\left((9\pi)/(10)\right))/((9\pi)/(10)) \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~ = (10)/(9\pi) \sin\left(\frac\pi{10}\right)`

`\displaystyle \prod_(k=1)^\infty \left(1 - (4^2)/(5^2(2k+1)^2)\right) = \prod_(k=1)^\infty \left(1 - \frac{4\left(\frac{2\pi}5\right)^2}{\pi^2(2k+1)^2}\right) \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = \prod_(k=2)^\infty \left(1 - \frac{4\left(\frac{2\pi}5\right)^2}{\pi^2(2k-1)^2}\right) \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = (5^2)/(3^2) \prod_(k=1)^\infty \left(1 - \frac{4\left(\frac{2\pi}5\right)^2}{\pi^2(2k-1)^2}\right) \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = \frac{25}9 \cos\left(\frac{2\pi}5\right)`

Then our sum is

`\displaystyle \prod_(k=0)^\infty (100k(k+1)+5^2)/(100k(k+1)+3^2) = \frac{\frac4{\pi^2}}{(100)/(9\pi^2) \sin^2\left(\frac\pi{10}\right)} \cdot \frac{25}9 \cos\left(\frac{2\pi}5\right) \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = \frac{2 \cos\left(\frac{2\pi}5\right)}{1-\cos\left(\frac\pi5\right)} \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = 1 + \sqrt5 \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = \boxed{2\phi}`

where we can use the approach I employed in another question of yours [28756378] to find the exact value of
`\cos\left(\frac\pi5\right)=\frac{1+\sqrt5}4`.