Convergence or divergence of $\sum\limits^{\infty}_{n=0} \frac{(2n+1)^{2}} {3^{n}(2n)!}$ 
Show if the series converges or diverges.
  $$\sum^{\infty}_{n=0}  \frac{(2n+1)^{2}} {3^{n}(2n)!}$$

Can someone please help with proving this? (I think it converges)
 A: $$\dfrac{(2n+1)^2}{(2n)!} = \dfrac{4n^2+4n+1}{(2n)!} = \dfrac{(2n)(2n-1)+6n+1}{(2n)!} = \dfrac1{(2n-2)!}+ 3 \cdot \dfrac1{(2n-1)!} + \dfrac1{(2n)!}$$
Now $$\sum_{n=0}^{\infty}\dfrac{(2n+1)^2}{(2n)!}x^n = \sum_{n=0}^{\infty}\dfrac{x^n}{(2n)!} + 3 \cdot\sum_{n=1}^{\infty}\dfrac{x^n}{(2n-1)!} + \sum_{n=1}^{\infty}\dfrac{x^n}{(2n-2)!}$$
\begin{align}
\sum_{n=0}^{\infty}\dfrac{x^n}{(2n)!} & = \dfrac{\exp(\sqrt{x})+\exp(-\sqrt{x})}2\\
\sum_{n=1}^{\infty}\dfrac{x^n}{(2n-1)!} & = \sqrt{x} \cdot \dfrac{\exp(\sqrt{x})-\exp(-\sqrt{x})}2\\
\sum_{n=1}^{\infty}\dfrac{x^n}{(2n-2)!} & = x \cdot \dfrac{\exp(\sqrt{x})+\exp(-\sqrt{x})}2
\end{align}
Hence,
$$\sum_{n=0}^{\infty} \dfrac{(2n+1)^2}{(2n)!} x^n = \dfrac{(x+\sqrt{x}+1) \exp(\sqrt{x})+(x-\sqrt{x}+1) \exp(-\sqrt{x})}2$$
A: Hint: Use the ration test (as suggested by @DonAntonio in the comment above)
$$
\left\lvert\frac{a_{n+1}}{a_n}\right\rvert = \frac{(2n+3)^23^n(2n)!}{3^{n+1}(2n + 2)!(2n+1)^2}
$$
Find the limit of this as $n\to \infty$. If the factorials are bothering your then think:
$$
\frac{(2n)!}{(2n+2)!} = \frac{2\cdot 3 \cdot 4 \cdot \dots \cdot (2n) }{2\cdot 3 \cdot 4 \cdot \dots \cdot (2n)\cdot (2n+1)\cdot (2n+2)}.
$$
