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**luzerne** Hi! I need to show that this double series diverge: $\displaystyle \sum\limits_{m=1}^{\infty}\sum\limits_{n=1}^{\inft y}\frac{1}{(m+n)^2}$

I first used the Integral Test on the inner summation (letting m fixed) with the improper integral $\displaystyle \int_{1}^{\infty}{\frac{1}{(m+x)^2}\,dx}$) and concluded it was finite, then computed a second improper integral with the integrand being my previous result (i.e $\displaystyle \int_{1}^{\infty}{\frac{1}{(1+x)}\,dx}$ ) and found this was diverging, and hence that the double series diverges as well.

But I think i get the right answer by accident: first I have never seen anywhere a "double improper integral test" and, more seriously, i know that the integral test tells us that if the integral converges then so does the series, but it doesn't tell that the series converges to the same value as the integral, yet i am using this value in my second integral...

I also need to show that $\displaystyle \sum\limits_{m=1}^{\infty}\sum\limits_{n=1}^{\inft y}\frac{1}{(m+n)^3}$ is finite and used the same method as above to find that it was indeed finite.

So is this approach justified? If not, how can we determine convergence and divergence for double summations?

You can indeed use a comparison with an integral; here is one way to procede:

Let $\displaystyle m\geq 1$. For any $\displaystyle n\geq 1$, we have (by monotonicity) $\displaystyle \frac{1}{(m+n)^2}\geq\int_n^{n+1}\frac{dx}{(m+x)^2 }$. Then,

$\displaystyle \sum_{n=1}^\infty \frac{1}{(m+n)^2}\geq \sum_{n=1}^\infty \int_n^{n+1}\frac{dx}{(m+x)^2}=\int_1^\infty \frac{dx}{(m+x)^2}=\frac{1}{m+1}$.

Since $\displaystyle \sum_m\frac{1}{m+1}$ diverges, the double series $\displaystyle \sum_m\sum_n\frac{1}{(m+n)^2}$ diverges.

In order to prove that $\displaystyle \sum_m\sum_n\frac{1}{(m+n)^3}$ converges, you can use the similar upper bound $\displaystyle \frac{1}{(m+n)^3}\leq\int_{n-1}^n\frac{dx}{(x+m)^3}$.

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There are other ways to prove this divergence. For instance (these are very similar to each other):

- for every $\displaystyle m\geq 1$, $\displaystyle \sum_{n=1}^\infty \frac{1}{(m+n)^2}=\sum_{p=m+1}^\infty \frac{1}{p^2}$ and (using a comparison with an integral) we have the well-worth-remembering estimate $\displaystyle \sum_{p=m+1}^\infty\frac{1}{p^2}\sim_m \frac{1}{m}$. Since $\displaystyle \sum_m\frac{1}{m}$ diverges, we conclude.

- for every $\displaystyle m\geq 1$, $\displaystyle \sum_{n=1}^\infty \frac{1}{(m+n)^2}=\sum_{p=m+1}^\infty \frac{1}{p^2}$, and, for $\displaystyle M\geq 1$:

$\displaystyle \sum_{m=1}^M \sum_{p=m+1}^\infty \frac{1}{p^2}=\sum_{m=1}^M\sum_{p=m+1}^M \frac{1}{p^2} + \sum_{m=1}^M \sum_{p=M+1}^\infty \frac{1}{p^2}$ $\displaystyle =\sum_{p=2}^M\sum_{m=1}^{p-1}\frac{1}{p^2}+M\sum_{p=M+1}^\infty\frac{1}{p^2}= \sum_{p=2}^M\frac{p-1}{p^2}+(\text{second term})$,

where the before-last equality is a "Fubini" rearrangement of the finite sum (interchanging sums). Then $\displaystyle M\to\infty$, the first term is a divergent series, and the second term is positive, hence the divergence.

- you may know that since the double series has positive terms, re-arranging the terms does not alter the convergence/divergence of the sum. Then we can group the terms where $\displaystyle m+n=p$. There are $\displaystyle p-1$ of them ($\displaystyle (1,p-1),(2,p-2),\ldots,(p-1,1)$), so that $\displaystyle \sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{(m+n)^2}=\sum_{p=2}^\infty \frac{p-1}{p^2}=+\infty$.

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You can prove a theorem of comparison with a double-integral. Suppose $\displaystyle fx,y)\mapsto f(x,y)$ is decresasing in both variables, and positive. Then for any $\displaystyle m,n\geq 1$, monotonicity gives:

$\displaystyle \int_m^{m+1}\int_n^{n+1} f(x,y) dy dx \leq f(m,n)\leq \int_{m-1}^m\int_{n-1}^n f(x,y) dy dx$.

Summing on $\displaystyle m,n$, we get:

$\displaystyle \int_1^\infty \int_1^\infty f(x,y)dy dx\leq \sum_{m=1}^\infty\sum_{n=1}^\infty f(m,n)\leq \int_2^\infty\int_2^\infty f(x,y)dy dx$.

Then if the left-hand side integral diverges, so does the double-sum, and if the right-hand side integral converges, so does the double-sum.