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Is the following series convergent or divergent?

$$\displaystyle{ \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{\sin(n-m)}{n^2+m^2}}$$

Even if it converge I do not know to prove it. However, for example, I know that $$\sum_n \frac{\sin n}{n}$$ converges. Then, why should not converge also this double series?

Thanks in advance.

Daniel W. Farlow
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Mark
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1 Answers1

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For fixed n, let $S(m,n) = \sum_{k=1}^{m}\sin (n-k).$ Then there exists $M$ such that $|S(m,n)|\le M$ for all $m,n.$ Summing by parts for fixed $n$ we find

$$\sum_{m=1}^{\infty}\frac{\sin(n-m)}{m^2+n^2} = \sum_{m=1}^{\infty}S_m[\frac{1}{m^2+n^2}- \frac{1}{(m+1)^2+n^2}]= \sum_{m=1}^{\infty}S_m\frac{2m+1}{(m^2+n^2)((m+1)^2 + n^2)}.$$

Slap absolute values everywhere in this last sum to see its absolute value is less than

$$M\sum_{m=1}^{\infty}\frac{3m}{(m^2+n^2)^2} =\frac{1}{n^2}\sum_{m=1}^{\infty}\frac{m/n}{((m/n)^2 + 1)^2}\frac{1}{n}.$$

As $n\to \infty,$ the last sum on $m$ tends to

$$\int_0^\infty \frac{x}{(x^2+1)^2}dx < \infty.$$ It follows that there is a positive constant $C$ such that

$$\left |\sum_{m=1}^{\infty}\frac{\sin(n-m)}{m^2+n^2} \right | \le \frac{C}{n^2}$$

for all $n.$ That implies the original double sum is convergent.

zhw.
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    Is $M$ depending on $n$ or not? – Elaqqad May 01 '15 at 19:35
  • It's independent of $n.$ Sum the sines as the imaginary part of a geometric sum of exponentials (the usual trick) to see this. – zhw. May 01 '15 at 19:40
  • @Elaqqad If we write $S(n,m) = \sum_{k = n-m}^{n-1} \sin k$, then multiply $\csc(1/2)\sin(1/2)$ and use the identity $\sin a \sin b = (1/2)[\cos(a-b) - \cos(a+b)]$, we obtain $$S(m,n) = (1/2)\csc(1/2)\sum_{k = n-m}^{n-1} [\cos(k - 1/2) - \cos(k + 1/2),]$$ which telescopes to $(1/2)\csc(1/2){\cos(n-m-1/2) - \cos(n+1/2)}$. This is bounded by $\csc(1/2)$, so we may choose $M = \csc(1/2)$, a number independent of $m$ and $n$. – kobe May 01 '15 at 19:54
  • The $S(n,m)$ I wrote was meant to be $S(m,n)$, sorry. :) – kobe May 01 '15 at 20:01