Check the convergence of series $$\sum _{n=1}^{\infty } \frac{\left(2 n^2-n+1\right)!}{3^{n^2+1}}$$
Trying to solve with sign d'Alembert, nothing comes out, and prevents the transformation of quadratic factorial reduction. Wolfram Alpha writes that decided by comparison, and I even had not come close to what can be compared.
 A: For a positive series to converge, the $n$-th term in the series must go to $0$ i.e. $\displaystyle\lim_{n\to\infty} \frac{(2n^2-n+1)!}{3^{n^2+1}}=0$.
But for $n\geq3$, we have $2n^2-n+1> n^2+1$. Hence $\displaystyle\frac{(2n^2-n+1)!}{3^{n^2+1}}=\frac{(2n^2-n+1)}{3}\frac{2n^2-n}{3}\cdots\frac{n^2-n}{3}\cdot{(n^2-n-1)!}>1$. We have shown that the $n$-th term doesn't go to 0 which implies that the series diverges.
A: Consider the $n$th entry $x_n=\dfrac{(a_n)!}{3^{b_n}}$ with $a_n=2n^2-n+1$ and $b_n=n^2+1$. The following facts hold for every $n\geqslant1$:
$$a_n\geqslant n^2\qquad a_{n+1}\geqslant a_n+4n\qquad b_{n+1}-b_n\leqslant 3n$$
The two first facts yield
$$
(a_{n+1})!=(a_n)!\prod_{k=1}^{a_{n+1}-a_n}(a_n+k)\geqslant(a_n)!\,(a_n)^{a_{n+1}-a_n}\geqslant(a_n)!\,n^{8n}.
$$
The last fact yields
$$
3^{b_{n+1}}\leqslant3^{b_n}\,3^{3n}.
$$
Thus,
$$
\frac{x_{n+1}}{x_n}\geqslant\frac{n^{8n}}{3^{3n}}=\left(\frac{n^8}{27}\right)^n\geqslant2,
$$
for every $n\geqslant2$. In particular, $x_n\to\infty$.
A: If $\displaystyle f(n)=2n^2-n+1,f(n+1)=2(n+1)^2-(n+1)+1$
So, $\displaystyle f(n+1)-f(n)=2(2n+1)-1=4n+1 $
If $\displaystyle g(n)=n^2+1, g(n+1)=(n+1)^2+1\implies g(n+1)-g(n)=2n+1$
So, if $\displaystyle T_n=\frac{\{f(n)\}!}{3^{g(n)}},$
$$\frac{T_{n+1}}{T_n}=\frac{\{f(n+1)\}!}{\{f(n)\}!}\cdot\frac1{3^{g(n+1)-g(n)}}=\frac{\{f(n+1)\}!}{\{f(n)\}!}\cdot\frac1{3^{2n+1}}$$
Observe that $\displaystyle\frac{\{f(n+1)\}!}{\{f(n)\}!}$ contains $4n+1$ multiplicands $2n^2-n+1+r, 1\le r\le 4n+1$
so is $\displaystyle O(n^{2(4n+1)})=O(n^{8n+2})$
$$\implies\lim_{n\to\infty}\frac{T_{n+1}}{T_n}\approx \lim_{n\to\infty}\frac{O(n^{8n+2})}{3^{2n+1}}=\lim_{n\to\infty}\left(\frac{n^8}{3^2}\right)^n\cdot\frac{n^2}3\to\infty$$
