Question

`lim_(n->oo)(1+1/n)^n =` ?

Answer 1

Let `a_n` and `b_n`, `n\inN`, are defined as:

`a_n=(1+1/n)^n`,

`b_n=(1+1/n)^(n+1) = a_n(1+1/n).`

It is obvious that:

`lim_(n->infty) a_n = (lim_(n->infty) b_n)/(lim_(n->infty) (1+1/n)) = lim_(n->infty) b_n.`

We'll show that sequence `b_n` is decreasing.

`b_(n-1)/b_n = (1+1/(n-1))^n/(1+1/n)^(n+1) = (n/(n-1))^n/((n+1)/n)^(n+1)=n^(2n+1)/((n-1)^n(n+1)^(n+1))`

`b_(n-1)/b_n = (n^2/(n^2-1))^n*n/(n+1) = (1+1/(n^2-1))^n*n/(n+1)`

Using Bernoulli's inequality `(1+x)^n > 1+nx` it follows:

`(1+1/(n^2-1))^n > 1+n/(n^2-1)` and:

`b_(n-1)/b_n > (1+n/(n^2-1))*n/(n+1) > (1+n/n^2)*n/(n+1)`

`b_(n-1)/b_n > (n(n+1))/n^2*n/(n+1) = 1`

So, the sequence `b_n` is decreasing.

Using Bernoulli's inequality again:

`b_n = (1+1/n)^(n+1) > 1+(n+1)/n = 1+1+1/n = 2+1/n > 2`

it follows that `b_n` has a lower bound. `b_n` is convergent.

Using inequality of arithmetic and geometric means:

`root(n+1)((1+1/n)^n*1) < (n(1+1/n)+1)/(n+1) = (n+2)/(n+1)`

it follows:

`(1+1/n)^n < ((n+2)/(n+1))^(n+1) = (1+1/(n+1))^(n+1),`

`a_n < a_(n+1)`

and `a_n` is increasing.

`a_n = (1+1/n)^n < (1+1/n)^(n+1) = b_n`

It follows that `a_n` has an upper bound (because `b_n` is decreasing and has a lower bound). Sequence `a_n` is increasing and has an upper bound so it's convergent.

Finally,

`e = lim_(n->infty) (1+1/n)^n.`

Answer 2

Let some value `x = lim_(n->oo) (1 + 1/n)^n`, where `n` is an arbitrary variable (i.e. does not have to be an integer!). Then the fastest way to proceed is as follows.

`lnx = ln(lim_(n->oo) (1+1/n)^n)`

`lnx = lim_(n->oo)(nln(1+1/n))`

`lnx = lim_(n->oo)(ln(1+1/n)/(1/n))`

By inspection, we've converted the right side into a `0/0` form. Therefore, we can use L'Hopital's rule.

`lnx = lim_(n->oo)((1/(1+1/n)*cancel(-1/n^2))/cancel(-1/n^2))`

`lnx = lim_(n->oo)(1/(1+1/n))`

wherein the `1/n` term vanishes as `n->oo`.

`lnx = lim_(nrarroo)(1/(1+0))`

`lnx=lim_(nrarroo)1`

`lnx=1`

Thus, we undo the natural logarithm to get:

`color(blue)(lim_(n->oo) (1+1/n)^n)`

`= e^(ln(lim_(n->oo) (1+1/n)^n))`

`= e^(lnx) = x = e^1 = color(blue)(e)`.

Answer 3

`(1+1/n)^n < (1+1/(n+1))^(n+1)` is increasing for `n->oo`

`(1+1/n)^n = 1+1/(1!)n/n+1/(2!)(n/n)((n-1)/n) + 1/(3!)(n/n)((n-1)/n)((n-2)/n)+cdots`

`((n-1)/n)(1+1/n)^n le 1-1+ (n-1)/n+1/(1!)(n-1)/n+1/(2!)((n-1)/n)^2+1/(3!)((n-1)/n)^3 + 1/(4!)((n-1)/n)^4+cdots +`

`((n-1)/n)(1+1/n)^n le (n-1)/n-1+sum_(k=0)^n1/(k!)((n-1)/n)^k`

so

`lim_(n->oo)((n-1)/n)(1+1/n)^n le lim_(n->oo)(-1/n)+lim_(n->oo)sum_(k=0)^n1/(k!)((n-1)/n)^k lelim_(n->oo)( -1/n)+lim_(n->oo)sum_(k=0)^n1/(k!)`

Finally

`lim_(n->oo)((n-1)/n)(1+1/n)^n = lim_(n->oo)(1+1/n)^n`

and

`lim_(n->oo)(1+1/n)^n le e`