Check if the following series converge $$\sum_{n=1}^{\infty}\frac{1 \cdot 4 \cdot 7 \cdot ... \cdot (3n+1)}{n^2 \cdot 3^n \cdot n!}$$
This is my solution:
Let's rewrite the question:
$$\sum_{n=1}^{\infty}\frac{(3n+1)!!!}{n^2 \cdot 3^n \cdot n!}$$
If we use D'Alembert's criterion:
$$\lim_{n \to \infty} \frac{\frac{(3n+4)(3n+1)!!!}{(n+1)^2\cdot3^n \cdot3 \cdot(n+1) \cdot n!}}{\frac{(3n+1)!!!}{n^2 \cdot 3^n \cdot n!}} = \lim_{n \to \infty} \frac{3n^3 + 4n^2}{3n^3 + 9n^2 + 9n+3} = 1$$
Which doesn't give an answer. However, if we use Raabe's criterion:
$$\lim_{n \to \infty} n\cdot(1-\frac{3n^3 + 4n^2}{3n^3 + 9n^2 + 9n+3}) = \frac{5}{3}$$
I calculated the limit correctly, I get the same result from WolframAlpha. Since the result is greater than 1, the series converge.
Is this correct?

Your proof is fine, as an alternative by limit comparison test we have
$$\frac{\frac{(3n+1)!!!}{n^2 \cdot 3^n \cdot n!}}{\frac1{n^\frac 43}}=\frac{(3n+1)!!!}{n^\frac23 \cdot 3^n \cdot n!} \to 0$$
indeed by Stirling's approximation we have
$$\frac{(3n+1)!!!}{n^\frac23 \cdot 3^n \cdot n!}=\frac{(n+\frac13)!}{n^\frac23 \cdot n!} \sim \frac{\sqrt {2\pi (n+\frac13)}\cdot \left(\frac {(n+\frac13)} e\right)^{(n+\frac13)}}{n^\frac23 \cdot \sqrt {2\pi n}\cdot \left(\frac n e\right)^n}=$$
$$=\sqrt{1+\frac1{3n}}\cdot \frac{(n+\frac13)^\frac13}{ e^\frac13\cdot n^\frac23}\cdot \left(1+\frac1{3n}\right)^n \to 0$$
therefore the given series converges by limit comparison test with $\sum \frac1{n^\frac43}$.
