Question: 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?

Answer: 
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: 

$$ \left(1+\frac1n\right)^n < e < \left(1+\frac1n\right)^{n+1}$$
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 $4\arctan \frac {1}{5} =$ $2\arctan \frac {5}{12}=$ $\arctan \frac {120}{119}=$ $\frac {\pi}{4}-\arctan \frac {1}{239}.$