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I need some urgent help in determining whether the following infinite series actually converges or not. Upon graphing, it does seem like so but I can't be sure. I'm not concerned with evaluating this sum at the moment. I tried some techniques on http://www.wikihow.com/Determine-Convergence-of-Infinite-Series but can't figure out for some and was indeterminable for others.

The Sum

The graph

Essentially if the series converges, that means there's an asymptote and that asymptote is the sum.

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Hint. One may use standard Taylor series expansions as $i \to \infty$, to obtain $$ \begin{align} &(2i+4)\tan\left(\frac{\pi }{2i+4}\right)+(2i+3)\tan\left(\frac{\pi }{2i+3}\right)=2 \pi +\frac{\pi ^3}{6 i^2}+O\left(\frac1{i^3}\right) \\\\ &\frac{\pi }{1-\frac{(i+2)\left(1-\sin\left(\frac{\pi (i+1)}{2i+4}\right)\right)^2}{2}}=\pi +O\left(\frac1{i^3}\right) \\\\ &\frac{\pi }{1-(2i+3)\left(\frac{1-\sin\left(\frac{\pi (2i+1)}{2(2i+3)}\right)}{1+\sin\left(\frac{\pi (2i+1)}{2(2i+3)}\right)}\right)^2}=\pi +O\left(\frac1{i^3}\right) \end{align} $$ giving, as $i \to \infty$, a general term such that $$ u_i=\frac{\pi ^3}{6 i^2}+O\left(\frac1{i^3}\right) $$ and the given series is convergent by the comparison test.

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  • $\begingroup$ Can you clarify how you found pi+O(1/i^3) for both expression? Sorry for a late question. $\endgroup$ – akira kato Nov 9 '16 at 7:01
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    $\begingroup$ @akirakato You can check the result step by step by plugging first "series of sin((\pi (i+1))/(2i+4)) for i=infinity" into WolframAlpha, then "series of (1-sin((\pi (i+1))/(2i+4)))^2 for i=infinity", and so on. $\endgroup$ – Olivier Oloa Nov 9 '16 at 9:56
  • $\begingroup$ By the way, I wanted to know if it converges because I gave up on evaluating that crazy sum. Do you know any potential way of evaluating it? $\endgroup$ – akira kato Nov 9 '16 at 12:18

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