Infinite sum of products on my mid-term This problem was on my calculus mid-term : Determine the convergence or divergence of the   series  $$
\sum\limits_{n = 1}^\infty  {\prod\limits_{k = 1}^n  {\frac{{4k - 3}}{{4k - 1}}} } 
$$
I tried everything, even the Wallis product formula, but nothing worked out. I'm pretty sure it diverges but I can't find a way to prove that the general term doesn't tend to 0.
 The following question was (I'm only including this because it involves roughly the same series) :
Determine the convergence or divergence of the alternating series: $$
\sum\limits_{n = 1}^\infty  {( - 1)^n \prod\limits_{k = 1}^n  {\frac{{4k - 3}}{{4k - 1}}} } 
$$

Set $$a_n:=\prod_{k=1}^n\frac{4k-3}{4k-1}.$$
Now
$$a_n\ge\frac{1}{3}\prod_{k=2}^n\frac{4k-5}{4k-1}=\frac{1}{4n-1},$$
which means that the first series diverges.
For the second series, consider $\log a_n$:
$$\log a_n=\sum_{k=1}^n\log\left(1-\frac{2}{4k-1}\right)
\le-\sum_{k=1}^n\frac{2}{4k-1}\to-\infty$$
as $n\to\infty$. This means $a_n\to 0$ as $n\to\infty$, and the alternating series converges relatively but (as we've seen from the first series) not absolutely.
