I had an alternative idea for an example that could be used to answer the 3.a part of the second part of the exam. \
Note that $\lim_{n\to\infty}(1+\frac{1}{n})^n=e$ and $lim_{n\to\infty}(\frac{2^{4n+1}n!^4}{(2n+1)(2n-1)!^2})=\pi$\
$$(x_n)=\frac{1}{2}\cdot((1+\frac{1}{n})^n+\frac{2^{4n+1}n!^4}{(2n+1)(2n-1)!^2})+\frac{1}{2}\cdot(1+\frac{1}{n})^n)\cdot(-1)^{n}+\frac{1}{2}\cdot(\frac{2^{4n+1}n!^4}{(2n+1)(2n-1)!^2})\cdot(-1)^{n+1}$$
Or written in a more elegant way
$$(x_n)=
\begin{array}{cc}
& \left\{
\begin{array}{cc}
(1+\frac{1}{n})^n & \text{n is even} \\
\frac{2^{4n+1}n!^4}{(2n+1)(2n-1)!^2} & \text{n is odd}
\end{array} \right.
\end{array}
$$
The only two converging subsequences of this sequence is the ones that converge towards $\pi$ and e and all elements of the sequence are rational numbers..
