Does #a_n = [n(n+1)]/(n+sqrt(n))# converge?

1 Answer
Apr 12, 2018

No

Explanation:

Because #a_"n"# is a functionnal sequence you only have to study limits ( if there's any) in #+oo# : #lim_"n->+oo"a_"n"= lim_"n->+oo"(n(n+1)) /(n+sqrt(n))#
#≈_"+oo"(n²)/n=n#

So
#lim_"n->+oo"a_"n"=+oo#
So no, #a_"n"# doesn't converge.
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