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 your approximations and using [Interval arithmetics](https://en.wikipedia.org/wiki/Interval_arithmetic), you just have to show that $5<e+\pi <6$.
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>).