## Question

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

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

## Answers

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

#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

#ln(3)/n -> 0# #ln(n) -> oo# #1-1/n -> 1#

#=>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#

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