Are These Two Lebesgue Integrals Different? I'm given a joint pdf $f: \mathbb{N}_1 \times \mathbb{N}_2 \rightarrow \mathbb{R}$ defined by $f(n,n) = 4^n -1 , f(n,n+1) = -2(4^n-1), f(n,m) = 0$ else alongside probability measure $\mu(\{n\}) = v(\{n\}) = \frac{1}{2^n} $. I then try to compute:
$$\int_\mathbb{N_1}\int_\mathbb{N_2}f(w_1,w_2)\mu(dx)v(dx) $$ 
as well as
$$ \int_\mathbb{N_2}\int_\mathbb{N_1}f(w_1,w_2)v(dx)\mu(dx)  $$
Now the way I solved each one gave me that they both equal $0$. This however I think is incorrect because I think the point of the problem is to show that we cannot use Fubini's theorem on these types of integrals?
The way I found that they both equal $0$ is by computing:
$$\int_\mathbb{N_1}\int_\mathbb{N_2}f(w_1,w_2)\mu(dx)v(dx) = \sum_{w_1=0}^\infty\sum_{w_2=0}^\infty f(w_1,w_2)P(w_1)P(w_2)=\frac{4-1}{4}-2\frac{4-1}{2\times 4}+\frac{4^2-1}{4^2}-2\frac{4^2-1}{2\times 4^2}+...=0$$ 
and then computing the other integral I get the same sum on the RHS which in turn gives $0$.
Is this the correct procedure, and do both yield $0$ as I found? Or am I just making a silly mistake in the double sum?
Thanks.
 A: So I'm answering my own question. The summation was incorrect and the correct way to do it is:
\begin{align*}
 \int_{\Omega_1}\int_{\Omega_2} f(\omega_1,\omega_2)\nu(d\omega_2)\mu(d\omega_1) &= \int_{\mathbb{N}_1}\int_{\mathbb{N}_2} f(n_1,n_2)\nu(dn_2)\mu(dn_1)\\
 &=\sum_{n_1=1}^{\infty} \sum_{n_2=1}^{\infty} f(n_1,n_2) \nu(dn_2)\mu(dn_1)\\
 &=\sum_{n_1=1}^{\infty} \frac{4^{n_1}-1}{2^{n_1}} + \frac{-2(4^{n_1}-1)}{2^{n_1 + 1}}\mu(dn_1)\\
 &=\sum_{n_1=1}^{\infty} \frac{4^{n_1}-1}{2^{n_1}} - \frac{(4^{n_1}-1)}{2^{n_1}}\mu(dn_1)\\
 &=\sum_{n_1=1}^{\infty} 0 \ \mu(dn_1)\\
 &=0\\
 \int_{\Omega_2}\int_{\Omega_1} f(\omega_1,\omega_2)\mu(d\omega_1)\nu(d\omega_2) &= \int_{\mathbb{N}_2}\int_{\mathbb{N}_1} f(n_1,n_2)\mu(d\omega_1)\nu(d\omega_2)\\
 &=\sum_{n_2=1}^{\infty} \sum_{n_1=1}^{\infty} f(n_1,n_2) \nu(dn_1)\mu(dn_2) \\
 &=\sum_{n_2=1}^{\infty} \left( \sum_{n_1=1}^{\infty} f(n_1,n_2) \mathbb{P}[N_1 = n_1] \right)\mu(dn_2)\ | \text{  Fix }n_2\\
 &=\sum_{n_2=1}^{\infty} \left( \mathbb{P}[N_1 = n_2]f(n_2,n_2) - \mathbb{P}[N_1 = n_2]f(n_2,n_2+1) \right)\mu(dn_2)\\
 &=\sum_{n_2=1}^{\infty} \left( \frac{4^{n_2}-1}{2^{n_2}} - \frac{2(4^{n_2}-1)}{2^{n_2}} \right)\mu(dn_2)\\
 &=\sum_{n_2=1}^{\infty} \frac{1 - 4^{n_2}}{2^{n_2}}\mu(dn_2)\\
 &\neq 0
 \end{align*}
