Does #a_n=n*{(3/n)^(1/n)} #converge? If so what is the limit?

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Answer 1 · 2016-12-02T11:13:32

#lim_(n->oo)a_n = oo#, thus #a_n# diverges.

#lim_(n->oo)a_n = lim_(n->oo)n(3/n)^(1/n)#

#=lim_(n->oo)e^ln(n(3/n)^(1/n))#

#=lim_(n->oo)e^(ln(n)+ln((3/n)^(1/n)))#

#=lim_(n->oo)e^(ln(n)+1/nln(3/n))#

#=e^(lim_(n->oo)ln(n)+ln(3/n)/n)#

Where the final step follows from the continuity of #f(x)=e^x#. Working on the new limit...

#lim_(n->oo)ln(n)+ln(3/n)/n = lim_(n->oo)ln(n)+(ln(3)-ln(n))/n#

#=lim_(n->oo)ln(3)/n+ln(n)(1-1/n)#

As #n->oo#, we have

#=>lim_(n->oo)ln(3)/n+ln(n)(1-1/n) = 0+oo*1 = oo#

Substituting this back in, we get

#lim_(n->oo)a_n=e^(lim_(n->oo)ln(n)+ln(3/n)/n)#

#=e^oo#

#=oo#

Thus #a_n# diverges as #n->oo#