With $a_{1}=1, a_{n}=n/(n-1),\text{ when }n > 1$ apply $c_{n}=\sqrt[n]{a_{1}\cdots a_{n}}$ which implies $\ln(c_{n})=\sum\ln(a_{n})/n$ to prove that $\sqrt[n]{n} \rightarrow 1$
So I know the easier way to get $\sqrt[n]{n} \rightarrow 1$, but I have to use apply $c_{n}$
So $\ln(c_{n})=\ln\left(\frac{(\frac{n}{n-1})}{n}\right)$ $\rightarrow$ $\frac{\ln(n)}{n} - \frac{\ln(n-1)}{n}$ $\rightarrow \ln(\sqrt[n]{n})-\ln(\sqrt[n]{n-1})$
Its at this point that I'm not sure what to continue doing. I could raise everything by e and then take the limit...