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

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tiburtius | High School Teacher | (Level 2) Educator

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`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)`.

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