Is there an elementary argument to show that $\prod_{n=0}^\infty \frac{2n+1}{2n+3}=0$ I managed to prove it by rewriting the partial products in terms of $n!$ and then applying Stirling's approximation. But I was wondering if there is a more elementary method to show that the limit is zero, without using such heavy machinery.
Also, is it true that in general $$\prod_{n=0}^\infty \frac{an+b}{an+c}=0, \qquad a>0\ \textrm{and}\ c>b$$
 A: The partial products are
$$ \prod_{n=0}^N\frac{2n+1}{2n+3}=\frac{1}{3}\cdot\frac{3}{5}\cdot\frac{5}{7}\cdots\frac{2N+1}{2N+3}=\frac{1}{2N+3} $$
hence
$$\prod_{n=0}^\infty\frac{2n+1}{2n+3}=\lim_{N\to\infty}\prod_{n=0}^N\frac{2n+1}{2n+3}=\lim_{N\to\infty}\frac{1}{2N+3}=0$$
The general product $\prod_{n=0}^{\infty}\frac{an+b}{an+c}$ requires a bit more work, because there's no reason why terms should cancel as they do in this case.
Taking a logarithm of the partial product, we get
$$ \log\frac{b}{c}+\sum_{n=1}^N\log\Big(\frac{1+\frac{b}{an}}{1+\frac{c}{an}}\Big) $$
and using $\log(1+x)=x+O(x^2)$ for small $x$, we get
$$ \log\Big(\frac{1+\frac{b}{an}}{1+\frac{c}{an}}\Big)=\log\Big(1+\frac{b}{an}\Big)-\log\Big(1+\frac{c}{an}\Big)=\frac{b-c}{an}+O(n^{-2}) $$
Therefore
$$ \log\frac{b}{c}+\sum_{n=1}^{\infty}\log\Big(\frac{1+\frac{b}{an}}{1+\frac{c}{an}}\Big)$$
diverges to $-\infty$, so the product converges to $0$.
A: You can write the infinite product as
$$\tag 1 \prod_{n=1}^\infty \left ( 1 -\frac{2}{2n+3}\right ).$$
Apply $\ln $ to get
$$\tag 2\sum_{n=1}^\infty \ln \left( 1 - \frac{2}{2n+3}\right ).$$
Now $\ln (1-u) < -u$ for $u \in (0,1),$ so $(2) \le \sum_{n=1}^\infty -2/(2n+3) = -\infty.$ Thus $(2)=-\infty,$ hence $(1)=0.$
Your more general product can be written as
$$\prod_{n=1}^\infty \left ( 1 -\frac{c-b}{an +c}\right )$$
and the same argument argument will work, nearly verbatim, to show the product converges to $0.$
A: For your first question, indeed there is an easy proof that 
$$
\lim_{k\to\infty} \prod_{n=0}^k \frac{2n+1}{2n+3} = 0
$$
Simply write $m=k-n$ in the product for a fixed $k$ to turn it into
$$
\prod_{m=0}^k \frac{2k-2m+1}{2k-2m+3} = \frac{2k+1}{2k+3} \cdot \frac{2k-1}{2k+1}
\cdot \frac{2k-3}{2k-1} \cdots \frac35 \cdot \frac13 = \frac{1}{2k+3}
$$
which indeed goes to zero as $k$ goes to infinity.
An identical technique works for your second problem as long as $c-b$ is a positive integer.  
If $c-b = q$ with $q\in \Bbb Q$ then multiply the numerator and denominator of the fractions by by the denominator of $q$ and the same technique works again.
To prove the limit is zero when $c$ and $b$ differ by an irrational quantity, you can start by proving that for a fixed $a$ and $b$ the product for any particular $k$ is strictly decreasing in $c$, and then bracket the product by two rational $c$ products.  Then the limit is bracketted by $0$ from above and $0$ from below. 
