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

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Answer 1 · 2017-02-07T13:41:25

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

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#