Limit Comparison test: Selection of auxiliary series How can we solve these questions using limit comparison test, specifically how can we choose $b_n$ for any series,
(The $n^{th}$ terms are given)
(1) $\dfrac{1.2.3.....n}{1.3.5....(2n-1)}$
(2) $\dfrac{(2n-1)}{n!}$
(3) $\dfrac{(n+1)(1)}{(n)(4^{n-1})}$

In the last question, I took $b_n$ as $\dfrac{1}{4^n}$ and then showed that $\displaystyle\lim_{n\to\infty}\dfrac{a_n}{b_n}$ is $4$. Therefore, both the series either converge or diverge. As $b_n$ is an infinite geometric series with sum equal to $\dfrac{4}{3}$ hence the given series also converges.

Is my procedure correct? 
The primary problem that I am facing while doing questions (1) and (2) pertains to the selection of $b_n$. 
Please help.
 A: For (b) use
$$0<\frac{2n-1}{n!}<\frac{2n}{n!}=\frac{2}{(n-1)!}<\frac{2}{n^2}$$
so compare with $\displaystyle\sum\frac{2}{n^2}$
For (a) Hint
$$\frac{1.2.3\cdots n}{1.3.5\cdots(2n-1)}=\frac{1.2.3\cdots n}{1.3.5\cdots(2n-1)}\frac{2.4.6\cdots(2n)}{2.4.6\cdots(2n)}=\frac{2^n(n!)^2}{(2n)!}$$
A: For the first series, we notice that
$$\lim_{n \to \infty} \frac{(1)(2)(3)\cdots(n)}{(1)(3)(5)\cdots(2n-1)}$$
$$=\lim_{n \to \infty} \frac{n!}{(2n-1)!!}$$
$$=\lim_{n \to \infty} \frac{2^n(n!)^2}{(2n)!}$$
$$=\lim_{n \to \infty} \frac{2^nn!}{(n+1)(n+2)\cdots(2n)}$$
$$=\lim_{n \to \infty} \frac{(1)(2)(3)\cdots(n)}{\frac{n+1}{2}\frac{n+2}{2}\frac{n+3}{2}\cdots\frac{n+n}{2}}$$
Clearly each term in the denominator will be greater than the associated term in the numerator (except for equality in the last term) for any finite $n$. We can now let $n \to \infty$ and conclude. Since the sum of the natural numbers is clearly unbounded, the partials sums of your first series must be unbounded and thus the series diverges. If you have to use Limit Comparison Test then I recommend first applying Stirling's Approximation to save you and me a headache.
Series B trivially converges. The numerator includes only half as many terms as the numerator, so the limit is easy. If you have to use Limit Comparison Test, we can easily let $b_n$ be $1$ over all the even factors in $n!$ multiplied together and conclude.
Series C seems a little messed up as-is (what is that factor of $1$ doing there??) but your logic seems correct.
