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I want to evaluate this alternating sum using few of standard summation $\sum_{n=0}^{+\infty}(-1)^n \tan(\frac{1}{n!})$ but i didn't succeed. This sum has alternating sign and it gives $0$ for some values of $n.$ Really the latter mixed me to know the exact sign of that sum however wolfram alpha assumed that converge for large $n$ to $\sim0.41\cdots$ , Now my question here is:

Question: What is the sign and exact value of this sum: $\sum_{n=0}^{+\infty}(-1)^n \tan(\frac{1}{n!})$ ?

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closed as off-topic by steven gregory, José Carlos Santos, Isaac Browne, Xander Henderson, JonMark Perry Jul 1 '18 at 13:10

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    $\begingroup$ I don't know why you'd think this was possible. Regardless, what's the definition of $n!$ for $n<0$? $\endgroup$ – David C. Ullrich Jun 29 '18 at 16:58
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    $\begingroup$ If you find the sum of the first $10$ or so terms and if you can then prove that the sum of the remaining terms is very small in absolute value, then that gives you the sign. When $x$ is close to $0$, as is $1/n!$ when $n$ is large, then the ratio of $\tan x$ to $x$ is close to $1. \qquad$ $\endgroup$ – Michael Hardy Jun 29 '18 at 17:09
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    $\begingroup$ You can also exploit the fact that $\lim\limits_{x\rightarrow0}\frac{\tan{x}}{x}=1$, thus $\tan{\frac{1}{n!}}\sim \frac{1}{n!}$. $\endgroup$ – rtybase Jun 29 '18 at 17:12
  • $\begingroup$ This can be resummed from n=2 -- the first two terms cancel. I think what results after changing order of summation might require acrobatics with the Kampé de Fériet function for a closed form. $\endgroup$ – graveolensa Jun 29 '18 at 17:45
  • $\begingroup$ @MichaelHardy The sign is positive for every Leibniz series with positive first term (and we can consider the series starting at $n=2$), because the sum of the first two terms is less than the sum. $\endgroup$ – egreg Jul 1 '18 at 13:05
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As said in comments, the best you can do is just to compute terms.

$$S_p=\sum_{n=0}^{p}(-1)^n \tan(\frac{1}{n!})$$ will give the following table $$\left( \begin{array}{cc} 0 & 1.55740772465490223051 \\ 1 & 0.00000000000000000000 \\ 2 & 0.54630248984379051326 \\ 3 & 0.37807527154154805200 \\ 4 & 0.41976606761920965972 \\ 5 & 0.41143254137928324024 \\ 6 & 0.41282143116123408937 \\ 7 & 0.41262301846021771374 \\ 8 & 0.41264782004752438635 \\ 9 & 0.41264506431560198078 \\ 10 & 0.41264533988879422065 \\ 11 & 0.41264531483668583521 \\ 12 & 0.41264531692436153399 \\ 13 & 0.41264531676377109562 \\ 14 & 0.41264531677524184122 \\ 15 & 0.41264531677447712485 \\ 16 & 0.41264531677452491962 \\ 17 & 0.41264531677452210817 \\ 18 & 0.41264531677452226436 \\ 19 & 0.41264531677452225614 \\ 20 & 0.41264531677452225655 \\ 21 & 0.41264531677452225653 \end{array} \right)$$ which not recognized by inverse symbolic calculators.

What would be interesting to know is : how many $p$ terms are to be added to get $k$ exact significant figures ? Since it is an alternating series, we need to solve $$\tan(\frac{1}{(p+1)!}) < 10^{-k}$$ and since $\tan(\epsilon)\approx \epsilon$, this reduces to $(p+1)! >10^k$.

If you look here, using $a=1$, @robjohn's fantastic approximation would write $${p\sim e\exp\left(\operatorname{W}\left(\frac{2k\log(10)-\log(2\pi)}{2e} \right)\right)-\frac32}$$ and then the results $$\left( \begin{array}{cc} k & \lceil p\rceil \\ 10 & 13 \\ 20 & 21 \\ 30 & 28 \\ 40 & 34 \\ 50 & 41 \\ 60 & 47 \\ 70 & 53 \\ 80 & 58 \\ 90 & 64 \\ 100 & 69 \\ 200 & 120 \\ 300 & 166 \\ 400 & 210 \\ 500 & 253 \\ 600 & 294 \\ 700 & 334 \\ 800 & 373 \\ 900 & 411 \\ 1000 & 449 \end{array} \right)$$

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