guys, as k-infinity, doesn't (1+1/k)^k tend toward 1, not e?

Question

guys, as k->infinity, doesn't (1+1/k)^k tend toward 1, not e?

anonymous · Answer

Whoops, nevermind... I was evaluating the limit of 1+1/k only, then raising it to the k power. My bad

phi · Answer

In case you are interested in this stuff, here is Khan's videos on compounding and e https://www.khanacademy.org/math/algebra2/exponential_and_logarithmic_func/continuous_compounding/v/introduction-to-compound-interest-and-e

amoodarya · Answer

$$(1+\frac{1}{k})^k=1+\left(\begin{matrix}k \ 1\end{matrix}\right)1^{k-1}(\frac{1}{k})^1+\left(\begin{matrix}k \ 2\end{matrix}\right)1^{k-2}(\frac{1}{k})^2+...=\1+\frac{k}{k}+\frac{k(k-1)}{2k^2}+...=\2+\frac{(k-1)}{2k}+...$$
obviously 
we have $$2<2+\frac{(k-1)}{2k}+...$$

amoodarya · Answer

if you try to find upper bound 
you will find $$2+\frac{(k-1)}{2k}+...<3$$

amoodarya · Answer

so 
$$2< (1+\frac{1}{k})^k<$$

amoodarya · Answer

in fact 
$$\lim_{k \rightarrow \infty} (1+\frac{ 1 }{ k })^k \rightarrow  e$$