Using the definition of convergence, how do you prove that the sequence #limit (sin n)/ (n) = 0# converges from n=1 to infinity?

Question

Scarlett Allison · Accepted Answer

undefined To prove that the sequence limit (sin n)/n = 0 converges from n = 1 to infinity using the definition of convergence, we need to show that for any ε > 0, there exists an N such that for all n ≥ N, |(sin n)/n - 0| < ε. Given ε > 0, we aim to find N such that |(sin n)/n - 0| < ε for all n ≥ N. Notice that |(sin n)/n| ≤ 1/n for all n ≥ 1 (as |sin n| ≤ 1 for all n). Let's choose N = 1/ε. Then for all n ≥ N, |(sin n)/n - 0| = |(sin n)/n| ≤ 1/n < 1/(1/ε) = ε. Therefore, for any ε > 0, we have found an N (specifically N = 1/ε) such that for all n ≥ N, |(sin n)/n - 0| < ε. Hence, by the definition of convergence, the sequence limit (sin n)/n = 0 converges from n = 1 to infinity.

Emilia Aguilar · Answer

Use the fact that, for #n > 1#, we have #abs sinn < 1#, so #abs (sinn/n) < 1/n# To show: #lim_(nrarroo)sin n/n = 0# We need to show that for any positive #epsilon#, there is a number #M#, such that if #n > M#, then #abs(sin n /n)< epsilon# Given #epsilon > 0#, Let #M# be an integer with #M > min{1, 1/epsilon}#. Note that #1/M < epsilon#. And if #n > M#, then #1/n < 1/M# and #abs(sin n / n -0) = abs (sin n)/n < 1/n < 1/M < epsilon#