Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.)

share|cite|improve this question
It's not true as stated. You probably meant to have a nondecreasing condition on $f$ that doesn't only apply on tuples of integers (otherwise you can take $f$ to be zero almost everywhere). Also, you seem to use $n_i$ to denote both integers and real variables, which isn't exactly wrong, but for example makes the phrase "for all $i$ and $n_i$" ambiguous. – Jonas Meyer Dec 23 '10 at 19:10
@Jonas: Thanks. I have clarified the condition a bit and replaced that "$+1$" with "$+\epsilon$" to show it is actually "monotonically decreasing in every direction". – kennytm Dec 23 '10 at 19:14
Thanks. It's still not true as stated. For example, let $N=2$ and define $f:[1,\infty)^2\to[0,\infty)$ by $f(x,y)=1$ if $x=1$, $f(x,y)=0$ otherwise. (It shouldn't be too hard to tweak this to a continuous example that shrinks rapidly to zero away from $x=1$.) – Jonas Meyer Dec 23 '10 at 19:34
@Jonas: Thanks. Someone please make that as an answer so I can accept it? – kennytm Dec 23 '10 at 20:58
up vote 6 down vote accepted

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

$$ \sum_{n_1=2}^\infty \cdots \sum_{n_N=2}^\infty f(n_1,\ldots,n_N) \leq \int_1^\infty \cdots \int_1^\infty f(x_1,\ldots,x_N) dx_1 \dotsm dx_N. $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.