I'm doing research on e and I don't understand how people know the function (1+1/n)^n is equal to a real number, and not 0, 1, or infinity. I'm in high school and don't really do a lot with limits, so when I saw the function (1+1/n)^n I thought it would tend to either 1 or infinity. My brain basically just put infinity for both n's even though I know that's not how it works.
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1$\left(1+\frac 1n\right)^n ={n \choose 0} + {n \choose 1} \frac1n+ {n \choose 2}\frac1{n^2}+\cdots + {n \choose n}\frac1{n^n} = 1 +1+ \frac12 \frac{n-1}{n}+\cdots $ using a binomial expansion so it is easy to show $\left(1+\frac 1n\right)^n\ge 2$ for $n\ge 1$ and not much harder to show $\left(1+\frac 1n\right)^n\le 3$. So if there is a limit as $n$ increases then it will be between $2$ and $3$. – Henry Mar 24 '25 at 09:41
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I totally understand the confusion... I was also confused when I was in high school. The problem involves a bit of understanding of how limits work, and how we can prove that sequences like this are "bounded" and not growing out of control. This is normally done in every first semester calculus course. We often need to use inequalities, like bounding the sequence between two values, to compute their limits if they exist. If you're interested, there's a really nice answer of the sequence being bounded here.
I´d suggest you to experiment with some of the great plotting tools like Desmos, to get a better intuition how these sequences look like. I hope this helps!
Idividedbyzero
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