Help with this irrationality proof

Question

I have a real number, that is quite messy so I'll just call it $x$. I want to prove it's irrational. It's a proof by contradiction. The contradiction will rise if I assume $x$ is a rational number $p/q$. I have arrived to the following equation: $$a_nx\pi+b_n=c_n$$ where $a_n$ and $b_n$ are integer sequences that depend on a natural number $n$. It's possible to show that $c_n < 1/n!$. The left hand side of the equation is positive. Notice that if I assume $x\pi$ to be a rational number $r/s$ the equation becomes $a_nr+sb_n=sc_n$. Since the left side is a positive integer and the right side is equal to $sc_n<s/n!$ this implies a positive integer less than $1$, so this means $x\pi$ can't be rational. However, I don't want to show that $x\pi$ is irrational, only $x$. So I'll assume that $x$ is a rational number $p/q$, then the equation becomes: $$a_np\pi+qb_n=qc_n $$ the right hand side goes to $0$, but the left hand side is not integer. I will change $\pi$ to its Leibniz series: $4s_n = \sum_{k=0}^n(-1)^k/(2k+1)$, where it gets close to $\pi$ as $n$ grows. So, if I choose a large $n$ I will have $$4a_nps_n+qb_n\approx qc_n $$ but the left hand side still is not integer because the denominator of $s_n$ are odd numbers up to $2n+1$. So I will multiple everything by the less common multiplier of all natural numbers up to $2n+1$, and I will call this $d_n$. So the equation becomes $$4d_na_nps_n+q d_nb_n\approx qd_nc_n $$ so I hope the left hand side has become integer. It is know that $d_n<e^{(2n+1)A}$ where $A$ is some positive real number. This means that the right hand side is $qd_nc_n<qe^{(2n+1)A}/n!$ which goes to $0$. But the left hand side is not equal to $qd_nc_n$, it's only an approximation. How do I proceed with this proof? Thanks.