Proving the divergence of arcsin(1/n)

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In summary, the given alternating series \sum (-1)^(n-1) * arcsin(1/n) is convergent, as proven by the alternating series test. To determine the type of convergence, the absolute value of the series, \sum arcsin(1/n), must be tested. The comparison test can be used, with the taylor series of arcsin as a comparison.
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mybluesock
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Homework Statement


Is [tex]\sum[/tex](-1)^(n-1)*arcsin(1/n) absolutely convergent, conditionally convergent, or divergent?


2. The attempt at a solution

The original function is alternating, so by the alternating series test, the function is convergent, because 0 < arcsin(1/(n+1)) <arcsin(1/n), and the limit of arcsin(1/n)=0.
So that rules out divergent. To determine whether the series is absolutely or conditionally convergent, you test the convergence of the absolute value of the series, which would be [tex]\sum[/tex] arcsin(1/n). However, I'm not sure what test to use now. Should I use the comparison test, and if so, what should I compare it to?
 
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i would attempt to use a comparison test...

considering the taylor series of arcsin will also be useful in showing the comparison is valid
 

1. What is the equation for arcsin(1/n)?

The equation for arcsin(1/n) is arcsin(1/n) = 1/n.

2. How do you prove the divergence of arcsin(1/n)?

To prove the divergence of arcsin(1/n), we can use the limit comparison test. This involves comparing the given series with another series whose divergence is already known. In this case, we can compare the series with the harmonic series, which is known to diverge. By taking the limit as n approaches infinity, we can show that the given series also diverges.

3. Can you provide an example of proving the divergence of arcsin(1/n)?

Sure, let's consider the series arcsin(1/n) = 1/n. We can compare this with the harmonic series 1/n. Taking the limit as n approaches infinity, we get:

lim (arcsin(1/n) / (1/n)) = lim (1 / n) / (1/n) = 1

Since the limit is greater than 0, we can conclude that the given series diverges.

4. What is the significance of proving the divergence of arcsin(1/n)?

Proving the divergence of arcsin(1/n) is important in determining the convergence or divergence of more complex series involving arcsin. It also adds to our understanding of the behavior of trigonometric functions and their corresponding series.

5. Are there any other methods to prove the divergence of arcsin(1/n)?

Yes, there are other methods such as the integral test, comparison test, and ratio test. However, in most cases, the limit comparison test is the easiest and most efficient method to prove the divergence of arcsin(1/n).

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