Question

test.

My thoughts : So for this one, I immediately thought of applying the test for divergence(which states if the limit of the nth term of the series, as n->infinity, is not equal to zero, then the series diverges)
Hence, we need to find lim n->infinity (sin(n)).


We know that the behaviour of the graph of sin(n) is periodic.

Answer 1

Hello, I have a question related to the convergence or divergence of an infinite series.
Question : Test the convergence/divergence of the series sin(n), using a suitable test.

My thoughts : So for this one, I immediately thought of applying the test for divergence(which states if the limit of the nth term of the series, as n->infinity, is not equal to zero, then the series diverges)
Hence, we need to find lim n->infinity (sin(n)).
But what is the value of the limit, lim n->infinity (sin(n))?
Does that exists or is it indeterminate?

We know that the behaviour of the graph of sin(n) is periodic.

So is sigma(sin(n)) divergent? What is the reason?

Or is there any other suitable test for testing this series for convergence or divergence?

Thank you

Answer 2

since the terms do not approach zero, it cannot converge.