Using the definition of convergence, how do you prove that the sequence #lim 2/(sqrt(n+3))=0# converges?

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Answer 1 · 2017-09-02T17:32:57

Given any real number #epsilon > 0#, we can choose an integer #N_epsilon# such that:

#(N_epsilon + 3)/4 > 1/epsilon^2#

Then for #n > N_epsilon# we have:

#2/sqrt(n+3) < 2/sqrt(N_epsilon+3)#

#2/sqrt(n+3) < 1/sqrt((N_epsilon+3)/4)#

#2/sqrt(n+3) < 1/sqrt(1/epsilon^2)#

#2/sqrt(n+3) < epsilon#

which proves the point.