If (An)=(1+1/n)^n then limit n-> infinity (An) is equal to: i) 0 ii) 1 iii) e iv) -1
`lim_(n->infty)(1+1/n)^n=e approx 2.718281828`
This is actually definition of irational number `e` which is the base of natural logarithm.
We can prove that this sequence is convergent by using the following theorem:
If sequence `a_n` of real numbers is monotone and bounded, then it is convergent.
By Bernoulli inequality
`(1+h)^alpha geq 1+alpha h`, for `h>0` and `alpha>1`
Now we put `alpha=(n+1)/n` and `h=1/(n+1)` to get
`(1+1/(n+1))^((n+1)/n)>1+(n+1)/n cdot 1/(n+1)=1+1/n`
Now we take `n`th power of both sides and to get
Hence, `a_n` monotonically increasing sequence.
Let's use Bernoulli inequality again
Again by takeing `n` th power we get
` ` `4>(1+1/n)^n=a_n`
which shows that `a_n` is bounded from above.
Thus `a_n` is convergent.
It can be shown that `lim_(n->infty)(1+a/(bn))^cn=e^((ac)/b)`.