Does the integral test work on higher dimensions? The integral test of convergence states that, if $f:[1,+\infty)\to[0,+\infty)$ is a monotonically decreasing nonnegative function, then the series $\sum_1^\infty f(n)$ converges iff $\int_1^\infty f(n) dn$ is finite.
Is the high-dimensional generalization also true? That is, given $f:[1,+\infty)^N \to[0,+\infty)$, and $f(\dotsc,n_i,\dotsc) \ge f(\dotsc,n_i+\epsilon,\dotsc)$ for all $1\le i\le N$ and $n_i\in[1,+\infty)$ and $\epsilon>0$, then the sum
$$ \sum_{n_1=1}^\infty  \cdots \sum_{n_N=1}^\infty f(n_1,\dotsc,n_N) $$
converges iff the multiple integral
$$ \int_1^\infty \cdots \int_1^\infty f(n_1,\dotsc,n_N) dn_1 \dotsm dn_N $$
is finite.
(This is just for checking if my answer over physics.SE is reasonable.)
 A: It isn't true in general, but the direction you used in your physics.SE answer is.  That is, if the sum converges, then the integral does too.  The decreasing hypothesis implies that the maximum value of $f$ on the cube $[n_1,n_1+1]\times\cdots\times[n_N,n_N+1]$ is $f(n_1,\ldots,n_N)$, so that the integral over that cube is less than or equal to $f(n_1,\ldots,n_N)$.  Adding up the integrals over all such cubes yields the result.
The problem with the other direction is that the function may drop off to zero in some directions rapidly enough to make the integral converge, while staying too big in another direction for the sum to converge.  For example, $N=2$, $f(x,y)=1$ if $x=1$, $f(x,y)=0$ otherwise.
With an additional "shift" you can go from integral convergence to sum convergence.  The decreasing hypothesis also implies that the minimum value of $f$ on the cube $[n_1,n_1+1]\times\cdots\times[n_N,n_N+1]$ is $f(n_1+1,\ldots,n_N+1)$, so that the integral over that cube is greater than or equal to $f(n_1+1,\ldots,n_N+1)$.  By adding up the integrals over all such cubes, this implies that 
$$
\sum_{n_1=2}^\infty  \cdots \sum_{n_N=2}^\infty f(n_1,\ldots,n_N)
\leq
\int_1^\infty \cdots \int_1^\infty f(x_1,\ldots,x_N) dx_1 \dotsm dx_N.
$$
