Why does $\left(1+ 1/k\right)^k$ converge to $e$ as $k \to\infty$? I came across this when learning about sequences and series. No one proved it to me tho! Is there a link anywhere? Or would it be beyond what I know? (First term of analysis)
 A: Of course, this depends on how you define $e$. Here is something that would convince a calculus student. 
Taking logs we have:
$\log((1 + \frac{1}{k})^k) = k \log(1 + \frac{1}{k})$ 
So we can take this limit as $k\rightarrow\infty$ using l'hospitals rule. 
$$ \lim_{k\rightarrow\infty} k \log(1 + \frac{1}{k}) = \lim_{k\rightarrow\infty} \frac{\log(1 + \frac{1}{k})}{\frac{1}{k}} = \lim_{k\rightarrow\infty} \frac{-\frac{1}{k^2 + k}}{-\frac{1}{k^2}} = \lim_{k\rightarrow\infty} \frac{k^2}{k^2 + k} = 1$$
Hence, using continuity of the function $e^x$, we conclude that 
$$\lim_{k\rightarrow\infty} (1 + \frac{1}{k})^k = e^1 = e$$
A: Usually Sometimes, $e$ is defined to be $\lim_{n \to \infty}\left(1+\frac 1n\right)^n$. What might need to be proved is that that limit exists.
Incidentally, this can be proved as well
$$\lim_{n \to \infty}\left(1+\frac 1n\right)^n = \sum_{n = 0}^\infty \frac{1}{n!}$$
I think a discussion of these things can be found in Rudin's PMA.
A: It's like calculating limit of function $\lim_{x\to \infty}(1+\frac1x)^x$
This is of the the form $1^{\infty}$.
So $$\lim_{x\to \infty}\left(1+\frac1x\right)^x = e^{\lim_{x\to\infty}((1+\frac1x)-1)(x)}$$
$$= e^1$$
$$= e$$
