<s>[INST] 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?
 [/INST] 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 $\sum \frac1{n^\frac43}$.
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