Is $\sum_{n=1}^{\infty}\frac{1}{n^{1+\left|{\cos n}\right|}}$ convergent?
Yes because $|\cos n|>0$ and $\frac{1}{n^ \alpha}$ is convergent for $\alpha>1$.
Is this a good way?
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1So, $\sum 1/ n^{1+1/n}$ converges? – David Mitra Dec 27 '14 at 20:13
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2No, this is not a good way. (And the question was already asked on the site.) – Did Dec 27 '14 at 20:14
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3Tastes a lot like http://math.stackexchange.com/questions/270064/does-the-series-sum-limits-n-1-infty-frac1n1-sinn-conve – Gabriel Romon Dec 27 '14 at 20:15
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1In order to apply Comparison directly, one would have to show that $|\cos n|\gt \beta$ for some fixed positive $\beta$. Unfortunately this is not true. – André Nicolas Dec 27 '14 at 20:15
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your alpha is not constant, so your series is not a p-series of Riemann – Idris Addou Dec 27 '14 at 20:15
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1@Imago what you're saying is wrong. – Gabriel Romon Dec 27 '14 at 20:15
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@ LeGrandDODOM: Ah, ty, I have just your post with the link and it really seems to be not that trivial. I had realized that you had a non constant alpha, but didn't think it through, if that would turn into a bigger problem. – Imago Dec 27 '14 at 20:17
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@LeGrandDODOM Thanks, this is the question I had in mind. – Did Dec 27 '14 at 20:24
2 Answers
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To be less subtle, you would need $|\cos n|\ge\alpha>0$ for some constant $\alpha$. Good luck with that!! :)
@DavidMitra's example is one of my favorites. Yes, $1+1/n\to 1$, but note that $$\lim_{n\to\infty}\frac{\frac1{n^{1+1/n}}}{\frac 1n}= 1,$$ so the limit comparison test says that the given series diverges.
Ted Shifrin
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1I disagree. One does not need such condition to have a convergent series. It is a sufficient condition, not a necessary one. – Jack D'Aurizio Dec 27 '14 at 21:03
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Of course, @Jack, but that was not the reasoning the OP suggested. – Ted Shifrin Dec 27 '14 at 21:05
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Not quite. To give a rigorous proof, you have to quantify how often $|\cos n|$ is close to zero. This depends on the irrationality measure of $\pi$. Since $\{n\pmod{\pi}\}_{n\in\mathbb{N}}$ is an equidistributed sequence, it is expected that $|\cos n|$ is, in average, $\frac{2}{\pi}$, giving a convergent series. However, this is not the case, as shown by the duplicate question's answers.
Jack D'Aurizio
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