How do I prove the following limit without using the derivative.
$$\lim_{n\to\infty}\big(1+1/n\big)^n$$
I have tried using the Binomial theorem but I haven't got too far. I have proved that the limit is between 2 and 3 and that it is convergent. Also if I logaritmate the whole equation I can't get anywhere without applying the derivative. And it doesn't get me anywhere if I try the Epsilon-delta proof.
$$\ln(y) = \ln\left(\lim_{n\to\infty}\big(1+1/n\big)^n\right)$$
On
Let $E(n)=(1+1/n)^n.$ Notice that $$\ln(E(n))=\frac{\ln(1+1/n)}{1/n}.$$ Therefore the limit as $n\to \infty$ of $\ln E(n))$ is $$ \lim_{1/n\to 0} \frac{\ln(1+1/n)-\ln(1)}{1/n} $$ which is the derivative of $\ln(x)$ near $1$ and therefore equals $1$. Using continuity of the natural logarithm, this shows that $$ \ln\lim_{n\to\infty}E(n)=1. $$
Therefore the limit must be $e$ where $e$ is the unique number whose natural log is $1$.
What you have written is often taken as the definition of $e$. You know that it converges, and the number it converges to is what we call $e$.
Another formula, useful in approximating $e$ (because it converges quickly), is this one: $$e = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \cdots = 1 + 1 + \frac{1}{2} + \frac{1}{6}+\cdots$$
To show that this is equal to what you've written, we should start with the binomial theorem. The $k$-th term of $(1+1/n)^n$ is ${n\choose k} n^{-k}$, or $\frac{n(n-1)\cdots(n-k+1)}{k! n^k} = \frac{1\cdot(1-1/n)\cdots(1-(k+1)/n)}{k!}$. If $k$ remains fixed, and $n\to \infty$, this is just $\frac{1}{k!}$.
This is not quite a proof, but it's a good start for comparing the two expressions. Once you know the above power series, you can compute that $e=2.7182818284\ldots$ or whatever you like.