How to derive the formula of the term in $ \sum_{n=1}^\infty \frac{1\cdot4\cdot \ldots \cdot(3n+1)}{(2n+1)!!} $ I stumbled upon the following exam problem
Test the convergence of the series:
$$ \sum_{n=1}^\infty \frac{1\cdot4\cdot \ldots \cdot(3n+1)}{(2n+1)!!} $$
So I figured every second factor of the numerator will cancel out with every factor in the denominator and that will give(actually not true)*
$$\sum_{n=1}^\infty4\cdot10\cdot16\cdot22...$$
In this particular case it might already be obvious that the series diverge but I wanted to derive an exact formula so I can prove the convergence or divergence with proper criteria/test. And after spending 10 minutes of trying to figure it out, I came up with the following formula $\ 2(2+3(n+1))=6n-2$. What turns out to be pretty easy to discover in this particular case if I did notice that those numbers were multiples of 6 - 2.
My question is, is there a known way to derive these formulas from infinite sums and infinite products? Or the deriving process becomes more easier just with time and practice?
I'm  pretty new to all this stuff, sorry if I'm missing the obvious with the question.
*As pointed out by @alex.jordan in the comments, I made a mistake there and the cancellation won't happen in the way I described. Nevertheless, it doesn't affect the essence of the question, therefore I will leave it unedited for now.
 A: Turning a comment thread into an answer:
Without explicitly finding a closed formula for the terms, you can still apply the Ratio Test. All terms are positive, so I will omit using absolute value that is in the more general form of the Ratio Test.
$$
\begin{align}
\frac{a_{n+1}}{a_n}
&=\frac{\frac{1\cdot4\cdot \ldots \cdot(3(n+1)+1)}{(2(n+1)+1)!!}}{\frac{1\cdot4\cdot \ldots \cdot(3n+1)}{(2n+1)!!}}\\
&=\frac{1\cdot4\cdot \ldots \cdot(3(n+1)+1)}{1\cdot4\cdot \ldots \cdot(3n+1)}\cdot\frac{(2n+1)!!}{(2(n+1)+1)!!}\\
&=\frac{\require{cancel}\cancel{1\cdot4\cdot \ldots \cdot(3n+1)}\cdot(3(n+1)+1)}{\cancel{1\cdot4\cdot \ldots \cdot(3n+1)}}\cdot\frac{(2n+1)!!}{(2n+3)!!}\\
&=(3n+4)\cdot\frac{(2n+1)!!}{(2n+3)\cdot(2n+1)!!}\\
&=\frac{3n+4}{2n+3}
\end{align}
$$
This expression goes to $\frac{3}{2}>1$ as $n\to\infty$, so by the Ratio Test, the original series diverges.
A: Consider
$$a_n=\frac{\prod_{k=0}^n (3k+1) } {(2n+1)!! }\qquad \text{and} \qquad S_p=\sum_{n=1}^p a_n$$ The first $S_p$'s are easy to compute; they generate the sequence
$$\left\{\frac{4}{3},\frac{16}{5},\frac{88}{15},\frac{1312}{135},\frac{2528}{165},\frac
   {34912}{1485},\frac{31648}{891},\frac{89504}{1683},\frac{1199776}{15147},\frac{5345248}{45441}\right\}$$ which is "almost" exponential.
Edit
Sooner or later, you will learn that
$$\sum_{n=0}^\infty a_n\,x^n=\, _2F_1\left(1,\frac{4}{3};\frac{3}{2};\frac{3 }{2}x\right)$$ which is the gaussian hypergeometric function which tends to $\infty$ when $x\to \frac 23$ from below.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
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\begin{align}
&\bbox[5px,#ffd]{{1\times 4 \times \cdots \times
\pars{3n + 1} \over \pars{2n + 1}!!}} =
{\prod_{k = 0}^{n}\pars{3k + 1} \over
\prod_{k = 0}^{n}\pars{2k + 1}} =
{3^{n + 1}\prod_{k = 0}^{n}\pars{k + 1/3} \over
2^{n + 1}\prod_{k = 0}^{n}\pars{k + 1/2}}
\\[5mm] = &\
\pars{3 \over 2}^{n + 1}\,
{\pars{1/3}^{\overline{n + 1}} \over
\pars{1/2}^{\overline{n + 1}}} =
\pars{3 \over 2}^{n + 1}\,
{\Gamma\pars{n + 4/3}/\Gamma\pars{1/3} \over
\Gamma\pars{n + 3/2}/\Gamma\pars{1/2}}
\\[5mm] = &\
{\root{\pi} \over \Gamma\pars{1/3}}\pars{3 \over 2}^{n + 1}\,
{\pars{n + 1/3}! \over \pars{n + 1/2}!}
\\[5mm] \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\, &
{\root{\pi} \over \Gamma\pars{1/3}}\pars{3 \over 2}^{n + 1}\,
{\root{2\pi}\pars{n + 1/3}^{\ n + 5/6}\expo{-n - 1/3} \over
\root{2\pi}\pars{n + 1/2}^{\ n + 1}\expo{-n - 1/2}}
\\[5mm]  \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\, &\
{\root{\pi} \over \Gamma\pars{1/3}}\pars{3 \over 2}^{n + 1}\,
{n^{n + 5/6}\,\bracks{1 + \pars{1/3}/n}^{\ n}\,\expo{-n - 1/3} \over
n^{n + 1}\,\bracks{1 + \pars{1/2}/n}^{\ n}\,\expo{-n - 1/2}}
\\[5mm] \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\, &
{\root{\pi} \over \Gamma\pars{1/3}}\,
{\pars{3/2}^{n + 1} \over n^{1/6}}
\,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\to}\,\,\,
\bbx{\large \infty} \\ &
\end{align}
