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sum_(k=1)^(oo)(1-cos ((1)/(k^(3))))

k=1(1cos1k3) \sum_{k=1}^{\infty}\left(1-\cos \frac{1}{k^{3}}\right)

Full solution

Q. k=1(1cos1k3) \sum_{k=1}^{\infty}\left(1-\cos \frac{1}{k^{3}}\right)
  1. Analyze Series Behavior: We are given the infinite series k=1(1cos(1k3))\sum_{k=1}^{\infty}(1-\cos (\frac{1}{k^{3}})). To find the sum of this series, we need to understand the behavior of the series as kk approaches infinity. The term (1cos(1k3))(1-\cos (\frac{1}{k^{3}})) approaches 00 as kk increases because the cosine of a very small number approaches 11. This suggests that the series might be convergent, but we need to analyze it further to be sure.
  2. Use Comparison Test: To analyze the convergence of the series, we can use the comparison test. We compare our series with a known convergent series. Since 1cos(x)1 - \cos(x) is approximately x22\frac{x^2}{2} for small xx, we can compare our series to the series k=1(12k6)\sum_{k=1}^{\infty}\left(\frac{1}{2k^6}\right), which is a pp-series with p=6p = 6. A pp-series is convergent if p > 1, so our comparison series is convergent.
  3. Show Inequality Holds: We need to show that for all kk, (1cos(1k3))(12k6)(1-\cos (\frac{1}{k^{3}})) \leq (\frac{1}{2k^6}). This is true for large kk because the second-order Taylor expansion of cos(x)\cos(x) around 00 is 1x22+O(x4)1 - \frac{x^2}{2} + O(x^4), and the term O(x4)O(x^4) becomes negligible for large kk. Therefore, our series is less than or equal to a convergent series term by term for large kk.
  4. Conclude Convergence: Since our series is less than a convergent series, by the comparison test, our series is also convergent. However, finding the exact sum of the series is not straightforward because the cosine function does not have a simple sum when taken over an infinite series like this one. We can conclude that the series converges, but we cannot find a closed-form for the sum.

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