This is not homework, just curiosity.
My question arose from the apparent absurdity that $\zeta(-1)=-\frac{1}{12}$, even though $\sum_{n=1}^\infty \frac{1}{n^z}$ only makes sense when $Re(z)>1$.
Let $f(x)=\sum_{n=0}^\infty x^n, \lvert{x}\rvert < 1$. It is known that $f(x)=\frac{1}{1-x}$. Let us now take the value of $f$ when $x=13$ (which also makes sense in the same way that taking $z=-1$ makes sense). We then have:
$$f(13) = \frac{1}{1-13}=-\frac{1}{12},$$ and consequently $$\sum_{n=0}^\infty 13^n=\sum_{n=1}^\infty n,$$ $$1+\sum_{n=1}^\infty 13^n=\sum_{n=1}^\infty n,$$ $$1+\frac{13+13^2+13^3+\cdots}{1+2+3+\cdots}=1,$$ and finally $$\frac{13+13^2+13^3+\cdots}{1+2+3+\cdots}=0.$$
Basically the idea is that in the case of $\zeta$, the LHS sum was divergent for the chosen parameter while the RHS was convergent, in the same way that in the case of $f$, the LHS diverged while the RHS was allowed to converge.
So, why does the above happen?
Your mistake is in going from $$1+\sum_{n=1}^\infty 13^n=\sum_{n=1}^\infty n$$ to $$1+\frac{13+13^2+13^3+\cdots}{1+2+3+\cdots}=1.$$ You divided the right side by $\sum_{n=1}^\infty n$, but on the left side you only divided the second term by $\sum_{n=1}^\infty n$, and you forgot to divide the $1$ by $\sum_{n=1}^\infty n$. If you do this step correctly, you get $$\frac1{\sum_{n=1}^\infty n}+\frac{\sum_{n=1}^\infty13^n}{\sum_{n=1}^\infty n}=1,$$ that is (correcting a big mistake in my original answer), $$-12+\frac{\sum_{n=1}^\infty13^n}{\sum_{n=1}^\infty n}=1.$$