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.
since the terms do not approach zero, it cannot converge.
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
To determine the behavior of the sequence (sin(n)), we need to find the limit as n approaches infinity.
Since the function sin(n) oscillates between -1 and 1 as n increases, it does not converge to a single value. Therefore, we can conclude that the limit does not exist.
To formally prove this, we can use the epsilon-N definition of a limit. We want to show that for any positive constant epsilon, there does not exist a natural number N such that if n is greater than or equal to N, then |sin(n)| < epsilon.
Since sin(n) oscillates indefinitely between -1 and 1, there will always be some values of n for which |sin(n)| is greater than any given epsilon. This means that we cannot find an N that satisfies the above condition for all epsilon.
In other words, the sequence (sin(n)) diverges because it does not converge to a specific value as n approaches infinity.