Asked by Robin
Determine whether the sequence is divergent or convergent.
a(sub n)=(−1)^n(sin(2/n))
a(sub n)=(−1)^n(sin(2/n))
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Answered by
Count Iblis
|a_n| = |sin(2/n)|
Since |sin(x)| <= |x|, we have:
|a_n| <= 2/n
The limit if |a_n| for n to infinity is thus 0, which then implies that the limit of a_n is zero.
You can easily make this rigorous, for every epsilon > 0, you can using the above inequality find an N such that for every n > N the absolute value of the difference of a_n and the limit is less than epsilon.
Since |sin(x)| <= |x|, we have:
|a_n| <= 2/n
The limit if |a_n| for n to infinity is thus 0, which then implies that the limit of a_n is zero.
You can easily make this rigorous, for every epsilon > 0, you can using the above inequality find an N such that for every n > N the absolute value of the difference of a_n and the limit is less than epsilon.
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