Use the Comparison Test to determine whether the series is convergent or divergent.
On top of the summation sign (∑) is infinity. Under the summation sign is n=1, and right next to it (to the right of ∑ sign) is cos^2n/n^2.
How would I compare them and come to the conclusion they are convergent with n grater or less than 1?
well sum of 1/n^2 converges, and cos^2(n)/n^2 <= 1/n^2
To determine whether the series (∑cos^2n/n^2) converges or diverges using the Comparison Test, we compare it to a known series with known convergence properties.
1. First, let's consider the fact that 0 ≤ cos^2n ≤ 1 for all values of n. This means that every term in the series (∑cos^2n/n^2) is between 0 and 1.
2. Now, we compare the series (∑cos^2n/n^2) to a known convergent series with positive terms such as the p-series (∑1/n^2).
3. By taking the ratio of the general terms of both series, we have:
lim (n→∞) (cos^2n/n^2) / (1/n^2).
4. Simplifying this expression, we get:
lim (n→∞) cos^2n.
5. Since the cosine function oscillates between -1 and 1, the limit of cos^2n doesn't exist.
6. However, despite the limit not existing, we can still use the Comparison Test. Since cos^2n is always positive and less than or equal to 1, we can say that the original series (∑cos^2n/n^2) is bounded between 0 and (∑1/n^2).
7. The series (∑1/n^2) is a known convergent series because it is a p-series with p = 2, and p > 1 ensures convergence.
8. Therefore, by the Comparison Test, if (∑1/n^2) converges, then (∑cos^2n/n^2) converges as well.
9. We know that the harmonic series of 1/n^2 converges (known proof), so the series (∑cos^2n/n^2) also converges.
In conclusion, the series (∑cos^2n/n^2) is convergent.