Is the sequence #a_n=(1+3/n)^(4n)# convergent or divergent?

Question

My issue is that the exponent is 4n, not n... I only know how to solve if it were just #n#

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Answer 1 · 2018-04-23T20:50:39

#"See explanation"#

#a_n = ((1+3/n)^4)^n#
#= (((1+3/n)^2)^2)^n#
#= ((1+6/n+9/n^2)^2)^n#
#= (1+36/n^2+81/n^4+12/n+18/n^2+108/n^3)^n#
#= (1+12/n+54/n^2+108/n^3+81/n^4)^n#

#"Note that you could more easily apply Euler's limit here :"#

#lim_{n->oo} (1+1/n)^n = e = 2.7182818....#

#=> lim_{n->oo} (1+3/n)^(12*n/3) = e^12 = 162754.79....#

#"So the sequence grows very big but not infinitely big, so it"#
#"converges."#