It was suggested Taylor series for that.

$e^1 = \sum_{k=0}^\infty \frac{1}{k!}$

I don't know how to prove the convergence of this series, so I tried to set the upper limit to 5 (I'm doing all this with a very simple calculator, it's basically by hand). Then $e \approx 2.71$

Since $4\arctan(1)= \pi$

$4\arctan(1)=4 \sum_{k=0}^\infty \frac{(-1)^k}{2k+1}$

Again doing a aproximation setting the upper limit to 5, $\pi\approx2.96$ (which I think it's pretty bad, but with my calculator it's the best I could do).

Then $e+\pi \approx 5.67$. But this only proves the approximation that I did is not integer, not the exactly value of $e+\pi$. Is there a way to prove that $e+\pi$ is not integer without relaying on approximations?

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    $\begingroup$ You don't have to use very accurate approximations... $\endgroup$ – Angina Seng Mar 2 '18 at 21:28
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    $\begingroup$ That series expression for $\pi$ converges very slowly. I would recommend Machin's $\pi = 16\arctan(1/5) - 4\arctan(1/239)$ instead. $\endgroup$ – Rolf Hoyer Mar 2 '18 at 21:43
  • $\begingroup$ @RolfHoyer I used the arctan series of $\pi$, because it's the only one I could come up with. I don't believe this "Machin's" series has a very trivial proof, so I rather avoid it, but I will look it up. Thanks! $\endgroup$ – Pinteco Mar 2 '18 at 21:48
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    $\begingroup$ Anon, to verify Machin's formula all you need is the formula for the tangent of a sum of two angles. Whenever I get a chance to teach power series I use this as an exercise. $\endgroup$ – Jyrki Lahtonen Mar 2 '18 at 22:04
  • $\begingroup$ @Anon: the proof only requires the formula for the tangent of a sum, which is used three times, and the knowledge of $\tan\pi/4=1$. One can consider it as elementary. $\endgroup$ – Yves Daoust Mar 3 '18 at 11:41

With your approximations and using Interval arithmetics, you just have to show that $5<e+\pi <6$.


With very crude approximations you can show that $ 5 \lt e +\pi \lt 6$, and the rest follows.

For crudely approximating $\pi$, we can use Taylor series expansion of $\tan^{-1}(x)$ Since it's an alternating series we can bound $\pi$ on both sides.

Since $\frac\pi6 = \tan^{-1}(\frac1{\sqrt{3}})$.

$$ 6(x - \frac{x^3}3) < \pi <6( x - \frac{x^3}3 + \frac{x^5}5) $$ $$ 3.07 < \frac{16 \sqrt 3}9 < \pi < \frac{82 \sqrt 3}{45} < 3.16 $$

For $e$ using the limit definition $\left(1+\frac1n\right)^n$ for two sided bounds:

$$ \left(1+\frac1n\right)^n < e < \left(1+\frac1n\right)^{n+1}$$

  • $\begingroup$ $3\frac {10}{71}<\pi<3\frac {1}{7}$ was given by Archimedes, obtained from from $48\sin \frac {\pi}{48}<\pi<48 \tan \frac {\pi}{48}.$ $\endgroup$ – DanielWainfleet Mar 3 '18 at 6:52

As @karakfa stated that you can prove $5\lt e+\pi\lt6$ using valid approximations. Better approximations would have been: $3\frac{10}{71}<\pi<3\frac{16}{113}$ for $\pi$ and $$\left(3+\sum_{i=2}^n \frac{-1}{i!(1-1)i}\right)\gt e \gt \sum_{i=0}^n \frac{1}{i!}$$ for $e$ (if $n$ is a finite natural number) as given in following web page: https://en.wikipedia.org/wiki/List_of_representations_of_e

You can set $n=5$ for $n$ in this approximation as you started.

The higher end of the approximation for $\pi$, $\pi \lt 3\frac{16}{113}$, is from A063674 (https://oeis.org/A063674).

  • $\begingroup$ @egreg: Thank you for your critical eye. I apologize for my mistake. It was corrected. $\endgroup$ – Mathew Mahindaratne Mar 3 '18 at 10:58

On calculating $\pi :$ If $\arctan u +\arctan v =\arctan 1=\frac {\pi}{4}$ we may estimate $\pi$ using $$\arctan x=x\sum_{n=0}^{\infty}\frac {(-x^2)^n}{2n+1}.$$

If $|ab|<1$ then $\arctan a + \arctan b =\arctan \frac {a+b}{1-ab}. $

$\frac {\pi}{4}=\arctan 1=\arctan \frac {1}{2}+\arctan \frac {1}{3}.$

$\arctan \frac {1}{2}=\arctan \frac {1}{3}+\arctan \frac {1}{7}.$

Hence $\frac {\pi}{4}=2\arctan \frac {1}{3}+\arctan \frac {1}{7}.$

Another one is $4\arctan \frac {1}{5} =$ $2\arctan \frac {5}{12}=$ $\arctan \frac {120}{119}=$ $\frac {\pi}{4}-\arctan \frac {1}{239}.$


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