<s>[INST] Show that $\sum^{\infty}_{n=1}{\frac{2(n+1)\cdot3^n}{\sqrt[3]{n!}}}$ is Convergent 
Show that $\sum^\infty_{n=1}\frac{n+1}{\sqrt[3]{n!}}\cdot(x^n+x^{-n})$ is uniformly convergent, when $-\frac{1}{3}\leq x\leq3$.

So I`m trying to show with Weierstrass M-test:
$|\frac{n+1}{\sqrt[3]{n!}}\cdot(x^n+x^{-n})|\leq|\frac{n+1}{\sqrt[3]{n!}}\cdot(3^n+3^{n})|={\frac{2(n+1)\cdot3^n}{\sqrt[3]{n!}}}=b_n$
In order to done, we need to show that the following 
$\sum^{\infty}_{n=1}{\frac{2(n+1)\cdot3^n}{\sqrt[3]{n!}}}$
is convergent, but it didn`t work for me with root\ratio tests.
Edit: It is working with ratio test and the solution for this is posted.
 [/INST] Ratio test:
$\lim_{n\to\infty} \frac{\frac{2(n+2)3^{n+1}}{\sqrt[3]{(n+1)!}}}{\frac{2(n+1)3^n}{\sqrt[3]{n!}}}$= $\lim_{n\to\infty}\frac{3\cdot(n+2)\cdot\sqrt[3]{n!}}{(n+1)\sqrt[3]{(n+1)!}}$=$\lim_{n\to\infty}\frac{3\cdot(n+2)}{\sqrt[3]{n+1}(n+1)}$=$3\cdot\lim_{n\to\infty}\sqrt[3]{\frac{(n+2)^3}{(n+1)^4}}$=$0 <1$
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