# I want to determine the convergence of the series to do so, consider that ( n

I want to determine the convergence of the series to do so, consider that
$\frac{\left(n+1{\right)}^{n}}{{3}^{n}n!}\le \frac{\left(n+1{\right)}^{n}}{{3}^{n}}$
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Tristin Case
So,
$\sum _{n=1}^{\mathrm{\infty }}\frac{\left(n+1{\right)}^{n}}{{3}^{n}}$
diverges, and, yes, you always have
$\frac{\left(n+1{\right)}^{n}}{{3}^{n}n!}⩽\frac{\left(n+1{\right)}^{n}}{{3}^{n}}.$
It doesn't follow from this that
$\sum _{n=1}^{\mathrm{\infty }}\frac{\left(n+1{\right)}^{n}}{{3}^{n}n!}$
diverges.
You can use the quotient test:
$\begin{array}{rl}\underset{n\to \mathrm{\infty }}{lim}\frac{\frac{\left(n+2{\right)}^{n+1}}{{3}^{n+1}\left(n+1\right)!}}{\frac{\left(n+1{\right)}^{n}}{{3}^{n}n!}}& =\underset{n\to \mathrm{\infty }}{lim}\frac{\left(n+2{\right)}^{n+1}}{3\left(n+1{\right)}^{n+1}}\\ & =\underset{n\to \mathrm{\infty }}{lim}\frac{1}{3}{\left(1+\frac{1}{n+1}\right)}^{n+1}\\ & =\frac{e}{3}\\ & <1,\end{array}$and therefore your series converges.