If (An)=(1+1/n)^n then limit n-> infinity (An) is equal to: 
i) 0 ii) 1 iii) e iv) -1

Expert Answers

An illustration of the letter 'A' in a speech bubbles

`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

`a_(n+1)=(1+1/(n+1))^(n+1)>(1+1/n)^n=a_n`

Hence, `a_n` monotonically increasing sequence.

Let's use Bernoulli inequality again

`4^(1/n)=2^(2/n)=1/2cdot2^(1+2/n)=1/2(1+1)^(1+2/n)>1/2(1+1+2/n)=1+1/n`

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)`.

See eNotes Ad-Free

Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.

Get 48 Hours Free Access
Approved by eNotes Editorial