p and k are real numbers. For which values of p and k does the following double series converge $$\sum_{n,m=1}^\infty \frac{1}{n^p + m^k}$$

I am trying to find a better (and quicker) way to solve this problem.

I'm trying to use RRL's hints (see below) to prove boundedness of the partial sums.

Edit: I was able to figure out the solution with the integral test method and think that I will move on to new problems now. But if anyone would like to post a solution that doesn't use the integral test and p-tests, that would be interesting to see - thanks in advance :-)


  • $\begingroup$ Can you use approximate by the double integral of say (1/(xp + yq)) - if this has a closed form for some p,q you'd probably have a good start. My guess is that p,q>1 just like the 1/n**p series. $\endgroup$ – user247608 Jul 10 '15 at 5:12


The summands are positive and decreasing with respect to each index when $p,k > 0$. The double series converges if the (monotone increasing) partial sums are bounded.

Note that with $f$ continuous and decreasing in each argument,

$$f(n,m) \leqslant \int_{n-1}^n \int_{m-1}^m f(x,y) \, dx \, dy.$$

Summing over $n,m = 2,3,\ldots,N$, we have

$$\sum_{n=2}^N \sum_{m=2}^N f(n,m) \leqslant \int_1^N \int_1^N f(x,y) \, dx \, dy.$$


$$\sum_{n=2}^N f(n,1) \leqslant \int_1^N f(x,1) \, dx, \\ \sum_{m=2}^N f(1,m) \leqslant \int_1^N f(1,y) \, dy. $$


$$\sum_{n=1}^N \sum_{m=1}^N f(n,m) \leqslant f(1,1) + \int_1^N f(x,1) \, dx + \int_1^N f(1,y) \, dy + \int_1^N \int_1^N f(x,y) \, dx \, dy.$$

If each integral is bounded for all $N \in \mathbb{N}$, then the double series converges.

Conditions for divergence can be found using the inequality

$$\sum_{n=1}^N \sum_{m=1}^N f(n,m) \geqslant\int_1^{N+1} \int_1^{N+1} f(x,y) \, dx \, dy.$$

  • $\begingroup$ The summands are positive and the sequence of partial sums is monotone increasing. It remains to show the boundedness of the sequence - thanks so much for this hint. I knew there'd be a simpler way to approach this problem. But, can you please elaborate on your first inequality? I'm not sure how you came up with it. Also, for the second and third terms (the single variable integrals), they each converge for p > 1 and k > 1 - these are just the usual p-integrals (like the p-series), when keeping the other argument fixed. However, we've already established that p, k must be greater than 1. $\endgroup$ – User001 Jul 10 '15 at 21:22
  • $\begingroup$ So, how does looking at these single-variable integrals again help us? Shouldn't we just look at the left-hand term less than or equal to the last term on the right hand side, the double integration in x and y? Thanks, @RRL $\endgroup$ – User001 Jul 10 '15 at 21:23
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    $\begingroup$ @LebronJames: You are essentially correct. I showed more detail in deriving the first inequality. In general, you just have to include checking that the single integrals converge to handle partial sums where $n =1$ and $m= 1$, $\endgroup$ – RRL Jul 10 '15 at 22:27
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    $\begingroup$ @LebronJames: Is it clear that the double series and integral converge and diverge together? That is what I showed. We certainly get divergence if either $p \leqslant 1$ or $k \leqslant 1$. For example with $f(x,y) = 1 / (x + y^2)$ the double integral diverges like $\log x$. Integration in closed form for arbitrary $p$ and $k$ results in the generalized hypergeometric function -- so you can look at the asymptotic behavior as the upper limt of integration tends to infinity. $\endgroup$ – RRL Jul 11 '15 at 0:29
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    $\begingroup$ @LebronJames: You're welcome. $\endgroup$ – RRL Jul 11 '15 at 4:36

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