Does this sequence diverge or converge?

Question

luigi0210 · Answer

$$\LARGE a_{n}=(1+\frac{2}{n})^n$$

anonymous · Answer

Converges to e^2

luigi0210 · Answer

Could you explain how you got that? I'm new to learning this stuff :P

ganeshie8 · Answer

substitute n = 2x,

$\LARGE a_{x}=(1+\frac{1}{x})^{2x} =  \left((1+\frac{1}{x})^{x}\right)^2  $

ganeshie8 · Answer

take the limit as x goes to infinity

luigi0210 · Answer

Thank you guys :)

ganeshie8 · Answer

np :) we use the known limit
$\large   \lim  \limits_{x \to \infty}  \left(1 + \frac{1}{x}\right)^x =   e$

ganeshie8 · Answer

the sequence converges if the limit $\lim \limits_{n \to \infty }a_n$ exists

luigi0210 · Answer

And since it got reduced to 
$$\Large \lim_{x \rightarrow \infty} (e)^2$$ 
it just became $\Large e^2$ right?

ganeshie8 · Answer

yes ! but it is illegal to show limit notation after taking the limit..

ganeshie8 · Answer

$$\large   \lim \limits_{x\to \infty }a_{x}=  \lim \limits_{x \to \infty  } \left((1+\frac{1}{x})^{x}\right)^2$$

$$\large     =\left(  \lim \limits_{x \to \infty  } (1+\frac{1}{x})^{x}\right)^2$$

$$\large     =\left(  e\right)^2$$

ganeshie8 · Answer

in the third step limit notation should not be present as the limit was already taken ^

luigi0210 · Answer

Oh, alright, thanks for the clarification gane :)

ganeshie8 · Answer

np :)

dan815 · Answer

boootifful