Why does the following series diverges? $$\sum_{i=1}^\infty \tan(\frac{\pi}{i+2})$$
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A related problem. – Mhenni Benghorbal Nov 21 '13 at 21:03
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It's easily solved by the comparison test to the harmonic series. – hardmath Nov 21 '13 at 22:44
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Since by the Taylor series we have $$\tan x\sim_0 x$$ so $$\tan\left(\frac{\pi}{i+2}\right)\sim_{i\to\infty}\frac{\pi}{i+2}\sim_{i\to\infty}\frac{\pi}{i}$$ hence your series is divergent.
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For $x$ small, $\tan(x) > x / 2$. So your series, after the first few terms, is term by term greater than $\sum_i \frac{\pi}{2(i+2)}$. Factor out a $\frac{\pi}{2}$, and you've got something that's greater than the tail of the harmonic series.
John Hughes
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I think "$x$ small" needs to be quantified... perhaps you can elaborate and give an argument that covers the full infinite spectrum of $x$? – abiessu Nov 21 '13 at 20:24
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1@abiessu I disagree; there is no need to quantify "small", since comparison with any tail of the harmonic series gives the desired result. – Andreas Blass Nov 21 '13 at 20:35
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@AndreasBlass: well, John has left out the comparison for $\tan(x)$ when $x$ is "not small", so I think it is still important... – abiessu Nov 21 '13 at 20:37
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@abiessu As I said, I disagree; I don't think the comparison for "not small" values is relevant to this question. – Andreas Blass Nov 21 '13 at 20:41
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@AndreasBlass: I respect your opinion, but I don't (and I expect the OP won't) understand why leaving out the "not small" portion is valid in this case. – abiessu Nov 21 '13 at 20:42