If n is very large (approaches infinity) the value of displaystyle{left({1}+frac{1}{{n}}
ight)}^{n}(1+
n
1
)
n
approaches e.
This is not an efficient way to find displaystyle{e}e. Even if we go out to n = 100,000, our value is only correct to the 4th decimal place.
displaystyle{e}approx{left[{left({1}+frac{1}{{n}}
ight)}^{n}
ight]}_{{{n}={100000}}}={2.718268237}e≈[(1+
n
1
)
n
]
n=100000
=2.718268237
Here's the graph demonstrating this expansion: