Proving the convergence of alternating series including pi product

Question

My class went over Alternating Series Test, which states that if series in the form of $\sum_{n=1}^{\infty}{\left(-1\right)^{n}b_{n}},\left({b_{n}}\ge0\right)$ is convergent if it satisfies both $$i.\text{ }b_{n}\ge b_{n+1} \forall n\ge 1$$ $$ii.\text{ }\lim_{n\to\infty} b_{n}=0$$ One of the practice question was to determine if the series $$\sum_{n=1}^{\infty}(-1)^{n+1}\left(\frac{n!}{1\cdot3\cdot5\cdot7\ldots(2n-1)}\right)$$ converges with Alternating Series Test.

With $b_{n}=\frac{n!}{1\cdot3\cdot5\cdot7\ldots(2n-1)}$, I was able to solve that the series satisfies i. However, I find it difficult to mathematically prove if $\lim_{n\to\infty}b_{n}=0$ is true.

I was able to use pi notation to simplify $b_{n}$ into $$b_{n}=\frac{n!}{1\cdot3\cdot5\cdot7\ldots(2n-1)}=\prod_{k=1}^{n}\frac{k}{2k-1}$$ Therefore now the question is $$\lim_{n\to\infty}\left(\prod_{k=1}^{n}\frac{k}{2k-1}\right)\overset{?}=0$$ to prove if the series meets criteria for Alternating Series Test.

Answer 1

Let $c_n=\log b_n=\log\prod_{k=1}^{n}\frac{k}{2k-1}=\sum_{k=1}^n\log(\frac{k}{2k-1})$.

Then $\lim_{n\to\infty}(c_n-c_{n-1})=\lim_{n\to\infty}\frac n{2n-1}=-\log2$.

So, $\lim_{n\to\infty}c_n=c_1+\lim_{n\to\infty}\sum_2^n(c_i-c_{i-1})=-\infty$.

So, $b_n=e^{c_n}\to0$.

Answer 2

Note that

$${b_{n+1}\over b_n}={n+1\over2n+1}\le{2\over3}$$

for all $n\ge1$, so, since $b_1=1$, we have

$$b_{n+1}={b_{n+1}\over b_n}\cdot{b_n\over b_{n-1}}\cdots{b_3\over b_2}\cdot{b_2\over b_1}\le\left(2\over3\right)^n\to0$$