Answer
Convergence
Work Step by Step
Given $$
\sum_{n=0}^{\infty} \frac{6^{n}}{(n+1)^{n}}
$$
By using the ratio test , since $a_n= \sum_{n=0}^{\infty} \frac{6^{n}}{(n+1)^{n}}$ and $a_{n+1}=\sum_{n=0}^{\infty} \frac{6^{n+1}}{(n+2)^{n+1}}$ , then
\begin{align*}
\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right|&=\lim _{n \rightarrow \infty}\left| \frac{6^{n+1}}{(n+2)^{n+1}} \frac{(n+1)^{n}}{6^{n}}\right|\\
&=\lim _{n \rightarrow \infty}\left| \frac{6(n+1)^{n} }{(n+2)^{n+1}} \right|\\
&=6\lim _{n \rightarrow \infty}\left| \frac{\left(n\left(1+\frac{1}{n}\right)\right)^n}{\left(n\left(1+\frac{2}{n}\right)\right)^{n\left(1+\frac{1}{n}\right)}}\right|\\
&=6\lim _{n \rightarrow \infty}\left|\frac{n^n\left(\frac{1}{n}+1\right)^n}{n^{n\left(\frac{1}{n}+1\right)}\left(\frac{2}{n}+1\right)^{n\left(\frac{1}{n}+1\right)}}\right|\\
&=6\cdot \frac{\lim _{n\to \infty \:}\left(\left(1+\frac{1}{n}\right)^n\right)}{\lim _{n\to \infty \:}\left(n\left(\frac{2}{n}+1\right)^{n\left(\frac{1}{n}+1\right)}\right)}\\
&=6\cdot \frac{e}{ \infty }\\
&=0<1
\end{align*}
Hence the given series is convergent.
