Problem with $ (1+\frac1n)^{n}$ I've noticed that if I consider $(1+\frac1n)^{n}$ and calculate its value when $n=10^3$ or $n=10^4$ I obtain a real number that is really close to $e$ but when I calculate it when $n=10^{11}$ or $n=10^{12}$ then I obtain $1$. I think it happens because for $n=10^{11}$ the  part $(1+\frac1n)$ is approximately $1$ then $1^n$ is $1$. But I also know that $\lim_{n\to+\infty}(1+\frac1n)^{n}=e$ so why for some value of $n$ I obtain $1$? Is it contradicting the limit? Thank you advance!!
 A: It is probably pa precision thing. Computers do not have infinite space and can therefore represent only a finite set of numbers exactly. Any operation you do on the numbers which results in a number that the computer cannot represent will result in rounding. 
This means that there exists some smallest number which is still greater than $1$ and is still representable by the computer, and if the result of $1+\frac1n$ is close enough to $1$, the computer will make an estimation that $1+\frac1n = 1$, therefore, $\left(1+\frac1n\right)^n = 1^n=1.$
A: The problem is your calculator or computer, not the mathematical limit. Because of the way floating point numbers work, they store a limited amount digits as well as the power. So even though $1\times 10^{-11}$ is valid, you can't store the number $1+10^{-11}$ accurately, since this would require too many digits of precision. Instead, the computer just rounds this number down to $1$, and you get your result.
A: In fact, the sequence $( 1 + \frac{1}{n})^{n}$ is an increasing sequence, which is bounded above by $e$, and $e$ is in fact the least upper bound for the sequence, hence is the limit.
The answer is that there is a fault in the computer's accuracy, there is no counterexample to the theory. If the computer contradicts the theory, then you can be sure that the fault is with the computer.
