2
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$$\sum_{n=1}^\infty \frac{(\ln n)^6}{\sqrt n}\cos \frac{\pi n} 6$$ Conditional convergence is easy : Let $\alpha = \frac{\pi n}{12}$, then $\cos2\alpha = 1 - 2\sin^2 \alpha$. $$\sum_{n=1}^\infty \frac{(\ln n)^6}{\sqrt n}\cos\frac{\pi n} n = \sum_{n=1}^\infty \frac{(\ln n)^6}{\sqrt n} - 2 \sum_{n=1}^\infty \frac{\sin^2 \alpha}{\sqrt n}$$ First series is divergent and second series is convergent, thus all series is divergent.

But I don't know what to do with absolute convergent. Could you give me any tips?

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    $\begingroup$ Sometimes $\ln^6 n$ means $\ln(\ln(\ln(\ln(\ln(\ln n)))))$ and sometimes it means $(\ln n)^6.$ I think it's better to avoid that notation except when you say which of those you intend. $\qquad$ $\endgroup$ – Michael Hardy Nov 4 '16 at 23:00
  • $\begingroup$ Do you mean $\cos(\pi n)/n$? $\endgroup$ – Clayton Nov 4 '16 at 23:02
  • $\begingroup$ I fixed a mistake. I mean cos(pi*n/6) $\endgroup$ – marka_17 Nov 4 '16 at 23:03
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    $\begingroup$ But $$\sum_{n = 1}^\infty \frac{\sin^2 \frac{\pi n}{12}}{\sqrt{n}}$$ diverges (and if you include the factor $(\ln n)^6$, it diverges a little faster). In fact, by Dirichlet's criterion the original series converges. $\endgroup$ – Daniel Fischer Nov 4 '16 at 23:09
  • $\begingroup$ @DanielFischer, i.e in original sereis cos is bounded and leftover part of term is monotonic and -> 0 ? $\endgroup$ – marka_17 Nov 4 '16 at 23:32

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