EDIT: as @saulspatz points out, I represented this entire thing incorrectly from the beginning so (all I'm looking for at this point is the correct way to set up) $\sum_{n=1}^{\infty}\frac{1\cdot8\cdot15\cdots(7n-6)}{7^nn!}$
EDIT: I've received the answer I'm looking for, @saulspatz spelled it out for me nicely.
I have to prove whether $$\sum_{n=1}^{\infty}\frac{1\cdot8\cdot15\cdots(7n-6)}{7^nn!}$$ converges or diverges (yes again, sorry to those who have to deal with my questions).
On
$$a_n=\frac{(7n-6)!}{7^nn!}\implies \log(a_n)=\log((7n-6)!)-n \log(7)-\log(n!)$$
Using Stirling approximation $$\log(a_n)=6 n (\log (n)-1+\log (7))-\left(6 \log (n)+\frac{11 \log (7)}{2}\right)+\frac{29}{14 n}+O\left(\frac{1}{n^2}\right)$$ Apply it a second time for $a_{n+1}$ and continue with Taylor series $$\log(a_{n+1})-\log(a_n)=6 \log (7 n)-\frac{3}{n}+O\left(\frac{1}{n^2}\right)$$ $$\frac{a_{n+1}}{a_n}=e^{\log(a_{n+1})-\log(a_n)}=(7n)^6+O\left(\frac{1}{n}\right)$$
The best way to approach these, I think, is to write out a few of the terms to get a feeling of what you are looking at. Here the first few terms are $$\frac17,\frac{8}{7^2\cdot2},\frac{8\cdot15}{7^3\cdot2\cdot3},\frac{8\cdot15\cdot22}{7^4\cdot2\cdot3\cdot4},\dots$$ Let's distribute the $7$'s, to make the terms in numerator and denominator more easily comparable. We have $$\frac17,\frac{8}{7\cdot14}\frac{8\cdot15}{7\cdot14\cdot21},\frac{8\cdot15\cdot22}{7\cdot14\cdot21\cdot28},\dots$$ Rewrite them once more as $$\frac17,\ \frac87\frac1{14},\ \frac87\frac{15}{14}\frac1{21},\ \frac87\frac{15}{14}\frac{22}{21}\frac1{28},\dots$$ and it's clear the terms are greater than those of the divergent series $\sum_{n=1}^\infty\frac1{7n}$ and the series diverges.
Now to write it up, I would just prove that $$\frac{1\cdot8\cdot15\cdots(7n-6)}{7^nn!}\geq\frac1{7n}$$