Question: I'm asked to show the convergence or divergence of the following series using the ratio test:

$$\sum\_{k=1}^{\infty} \frac{1\cdot 4\cdot 7\cdot\cdot\cdot (3k-2)}{2\cdot 4\cdot 6\cdot\cdot\cdot (2k)}$$

What does the "$\cdot\cdot\cdot$" represent, and how would I go about applying the ratio test to such a series (which is, of course, 'trivial' for something less.. obscure(?)) ?

Answer: 
As far as I know there is no analytical expression, but using the remainder formula for the Taylor expansion of the exponential,

$$e-\sum\_{k=0}^n\frac1{k!}=\frac c{(n+1)!}$$ where $1<c<e$. For $n=100$ or $n=1000$, this is microscopic.

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**Update:**

From the link given by Raymond, we have

$$c=(n+1){\{e\,n!\}}=1+\frac1{n+2}+\frac1{(n+2)(n+3)}+\cdots<1+\frac1{n+1},$$ which obviously tends to $1$.
I am not sure I understand your question, but finite sums are often referred to as [partial sums](https://mathworld.wolfram.com/PartialSum.html).

One can obtain a partial sum by simply setting a finite number as the upper limit of the sum (compared to infinity).

This way, one merely iterates through the appropriate values of $x$ (in this case) to obtain the partial sum (or finite series)

For example:

$\sum\_{x=1}^3 \frac{1}{x!} = \frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}=1+\frac{1}{2}+\frac{1}{6}=1+\frac{2}{3}$

This can be generalized as:

$\sum\_{x=n}^m f(x) = f(n)+f(n+1)+f(n+2)...+f(m)$

where $f$ is an function from the integers to the reals.