#lim_(n->oo)n^2(a^(1/n)+a^(-1/n)-2)#?

Question

450 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-05T11:42:57

#(log_e a)^2#

With #a > 0#, making #a=e^lambda# we have

#n^2(a^(1/n)+a^(-1/n)-2) = 2lambda^2(n/lambda)^2(cosh(lambda/n)-1)=#
#=2lambda^2((cosh(lambda/n)-1)/(lambda/n)^2)# Making now

#x=lambda/n# we have

#lim_(n->oo)n^2(a^(1/n)+a^(-1/n)-2)=lim_(x->0)2lambda^2((cosh(x)-1)/x^2)#

but #cosh(x) = 1 +x^2/(2!)+x^4/(4!)+ cdots #

then

#lim_(n->oo)n^2(a^(1/n)+a^(-1/n)-2) = lambda^2#

but #lambda = log_e a# then

#lim_(n->oo)n^2(a^(1/n)+a^(-1/n)-2) =(log_e a)^2#