## Precalculus (6th Edition) Blitzer

$\approx 2.7183$
We know that $a_n=1+(\dfrac{1}{n})^n$ Here, we have $a_{10}=1+(\dfrac{1}{10})^{10}=2.5937$ $a_{100}=1+(\dfrac{1}{100})^{100}=2.7048$ $a_{1000}=1+(\dfrac{1}{1000})^{1000}=2.7169$ $a_{10000}=1+(\dfrac{1}{10000})^{10000}=2.7181$ ..... and so on Hence, we can see that as the value of n increases, the value approaches to $2.7183$