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

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Answer 1 · 2015-10-30T14:42:42

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#