Question

How do you determine whether the sequence `a_n=((n-1)/n)^n` converges, if so how do you find the limit?

Answer 1

`lim_(n->oo) a_n =1/e`

Explanation:

We have:

`a_n =((n-1)/n)^n`

this is a positive number so we can take the logarithm:

`ln a_n = ln ((n-1)/n)^n = nln ((n-1)/n)`

Now if:

`lim_(x->oo) xln((x-1)/x)`

exists is must be the same as:

`lim_(n->oo) ln a_n = lim_(n->oo) nln ((n-1)/n)`

This limit in the indeterminate form `0*oo` so we have to transform it to apply l'Hospital's rule:

`lim_(x->oo) xln((x-1)/x) = lim_(x->oo) (ln((x-1)/x) )/(1/x) = lim_(x->oo) (d/dx ln((x-1)/x) )/(d/dx 1/x)`

Now:

`d/dx ln((x-1)/x) = d/dx (ln(x-1) - lnx) =1/(x-1) -1/x = 1/(x(x-1))`

`d/dx 1/x = -1/x^2`

so:

`lim_(x->oo) xln((x-1)/x) = lim_(x->oo) (1/(x(x-1)))/(-1/x^2) = lim_(x->oo) -x^2/(x(x-1)) = -1`

and we can conclude that:

`lim_(n->oo) ln a_n = -1`

But since `ln x` is a continuous function:

`lim_(n->oo) ln a_n = ln ( lim_(n->oo) a_n) = -1`

which implies:

`lim_(n->oo) a_n =1/e`