Show that $a_n=(n+1)^{\frac{1}{\ln(n+1)}} \to e$ when $n \to \infty$ Show that $a_n=(n+1)^{\frac{1}{\ln(n+1)}} \to e$ when $n \to \infty$ Can you give me a hint? I can't use L'Hopital. Thanks.
$${{(n+1)}^{\frac{1}{\ln (n+1)}}}={{(n+1)}^{\frac{\ln e}{\ln (n+1)}}}={{(n+1)}^{\log _{n+1}^{e}}}={{e}^{\log _{n+1}^{n+1}}}=e$$
Hint 1: It's a trick question. The solution is really, really, really easy. 

Hint 2: Take a spreadsheet program and calculate the values for n = 1 to 10.