$$\frac{e^{-2 i} (-1)^k \left(2 e^{2 i} \Gamma (k+1)+\Gamma (k+1,-2 i)+e^{4 i} \Gamma (k+1,2 i)\right)}{4 \Gamma (k+1)}\to k\to 1,2,(3)\text{..}\to \left\{1,-1,0,0,\frac{1}{3},-\frac{1}{3},\frac{13}{45},-\frac{13}{45},\frac{92}{315},-\frac{92}{315},\frac{4138}{14175}\text{...}\right\}\to \cos ^2(1)$$ the sequence of rational Taylor coefficient series tend to $\cos(1)^2$ as $k\to\infty$ but never reach $\cos(1)^2$ so it is irrational?

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    $\begingroup$ I can't follow what you're doing; your notation is highly nonstandard. But $\cos(1)^2$ is not only irrational but transcendental. $\endgroup$ – Henning Makholm Jan 16 '15 at 18:12
  • $\begingroup$ Look at the Taylor expansion of say $\frac{1}{1-x}$ for $x=1/2$. The sequence of Taylor approximations consists of rationals, and the limit is rational. $\endgroup$ – André Nicolas Jan 16 '15 at 18:16
  • $\begingroup$ Aside: $( \frac{1}{10^k} ) = ( \frac{1}{10}, \frac{1}{100}, \frac{1}{1000}, \frac{1}{10000}, \ldots )$ tends to $0$ as $k \to \infty$ without ever reaching $0$, but $0$ manages to be rational anyways. $\endgroup$ – Hurkyl Jan 16 '15 at 18:17
  • $\begingroup$ $$\frac{1}{2} + \frac{1}{2}(\cos(2)) = 0.29192658172... $$ and $$\cos^2(1)$$ is a transcendental number $\endgroup$ – Irrational Person Jan 16 '15 at 18:23

The argument with a sequence of rationals converging to the number in question shows nothing. For example, consider $\frac{n}{n+1}\to 1$.

But assume $\cos^2 1$ is rational. Then $\cos 1$ and $\sin 1$ and $\cos 1+i\sin 1=e^i$ are algebraic. But by Gelfond-Schneider, $e^i$ is transcendental.

  • $\begingroup$ But by Gelfond-Schneider, $e^i$ is transcendental - e is not algebraic. $\endgroup$ – Lucian Jan 16 '15 at 19:02
  • $\begingroup$ Why does $\cos^2(1)$ being rational imply that $\cos(1)$ is rational? $\endgroup$ – marty cohen Jan 16 '15 at 19:29

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