Prove that the series ∞Xn=1sin nn2is convergent
Question
Prove that the series ∞Xn=1sin nn2is convergent
Solution
The series you're referring to is ∑(n=1 to ∞) sin(n)/n^2.
To prove that this series is convergent, we can use the comparison test. The comparison test states that if 0 ≤ a_n ≤ b_n for all n and the series ∑b_n is convergent, then the series ∑a_n is also convergent.
In this case, we know that -1 ≤ sin(n) ≤ 1 for all n. Therefore, 0 ≤ |sin(n)/n^2| ≤ 1/n^2.
The series ∑(n=1 to ∞) 1/n^2 is a p-series with p = 2, which is greater than 1. Therefore, by the p-series test, ∑(n=1 to ∞) 1/n^2 is convergent.
By the comparison test, since 0 ≤ |sin(n)/n^2| ≤ 1/n^2 and ∑(n=1 to ∞) 1/n^2 is convergent, the series ∑(n=1 to ∞) sin(n)/n^2 is also convergent.
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