does this trigonometric series vanish?

Question

I have a question regarding a trigonometric series:

Shouldn't

$$\sum_{n,m}^\infty s^{n+m}sin\left((n-m)\phi\right)=0$$

where $s \in \mathbb{R}$ because the sinusoidal function is antisymmetric in $n$ and $m$ and vanishes for $n=m$?

Answer 1

$$\begin{align} & &\sum_{n,m}^\infty s^{n+m}\sin\left((n-m)\phi\right)\\&=& \text{Im}\sum_{n,m}^\infty s^{n+m}\text{e}^{(n-m)\text{i}\phi}\\&=& \text{Im}\sum_{n}^\infty (s\text{e}^{\text{i}\phi})^n \sum_{m}^\infty (s\text{e}^{-\text{i}\phi})^m\\&=& \text{Im}\frac{1}{1-s\text{e}^{\text{i}\phi}}\cdot\frac{1}{1-s\text{e}^{-\text{i}\phi}}\\ &=&\text{Im}\frac{1}{1-2\cos\phi s +s^2}\\ &=&0. \end{align} $$

Answer 2

You need to have absolute convergence to rearrange the serie, or make the summation in the order you wish.

Let examine for instance only the following simpler serie : $\sum\limits_{n}^{\infty} (-1)^n$.

The general term does not go to $0$ so the serie is divergent, it means you have no right to do the following thing :

$\sum\limits_{n=0}^{\infty}(-1)^n=\underbrace{1-1}_0+\underbrace{1-1}_0+\underbrace{1-1}_{0}+...=\sum\limits_{m=0}^\infty\big((-1)^{2m}+(-1)^{2m+1}\big)=\sum\limits_{m=0}^\infty(1-1)=\sum\limits_{m=0}^\infty 0=0$

Because then, one can also do this: $\sum\limits_{n=0}^{\infty}(-1)^n=1+\underbrace{-1+1}_0+\underbrace{-1+1}_0+...=1$

I have obtained $2$ different values for the sum of this serie, so the original expression $\sum\limits_{n}^{\infty} (-1)^n$ is undefined (or divergent).

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The same arise for double summation like yours, let's consider $\sum\limits_{n,m}^{\infty}(n-m)$

It has no meaning to say that:

$\sum\limits_{n,m}^{\infty}(n-m)=\sum\limits_{k=0}^{\infty}\sum\limits_{n+m= k}(n-m)=\sum\limits_{k=0}^{\infty}0=0$

The above is equivalent to doing this :

$(0)+\underbrace{-1+1}_{0}+\underbrace{-2+0+2}_{0}+\underbrace{-3-1+1+3}_{0}+\underbrace{-4-2+0+2+4}_{0}+\underbrace{-5-3-1+1+3+5}_{0}+...=0\tag{E}$

But I could also do this :

$(0-1+1-2+0)+\underbrace{2-3-1+1}_{-1}+\underbrace{3-4-2+0+2}_{-1}+\underbrace{4-5-3-1+1+3}_{-1}+...=-\infty$

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Let's now have $a(s,\phi,m,n)=s^{n+m}\,\sin((n-m)\phi)$

If $|s|<1$ then $|a(s,\phi,n,m)\le|s|^{n+m}\tag{*}$

which is a term of a congergent geometric serie, so the original serie is absolutely convergent.

Note that $a(s,k\pi,n,m)=0$ so in this case the serie also has sense, but in general for any $\phi$ and $|s|\ge 1$ then $a(s,\phi,n,m)\nrightarrow 0$ so the serie is divergent.

When we place ourselves in the absolute congence case, then both methods by @ancientmathematican @vrugtehagel have meaning because the intermediate series they are using are all convergent series.

We can also show that some partial series are convergent to zero.

Partial series are finite sums, so we can rearrange in any way we wish.

In particular $\sum\limits_{m=0}^{N}\sum\limits_{n=0}^{N}a(s,\phi,m,n)=0$ because terms are telescoping since it is symetric in $n,m$.

Or you could also consider $\sum\limits_{k=0}^{K}\sum\limits_{m+n=k}a(s,\phi,m,n)=0$ according to $(E)$ rearrangement.

Since it is absolutely convergent, the remainders are going to $0$ in both cases while $(n,m)\to(\infty,\infty)$, this is why we can examine various kind of partial series.

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To show absolute convergence of $(*)$ for instance, we would have to show that partial serie is convergent.

That is $S(N,M)=\sum\limits_{n=0}^{N}\sum\limits_{m=0}^{M}|s|^{n+m}$ is convergent to a single limit when $(N,M)\to(\infty,\infty)$.

This time we cannot make assumptions like $M=N$ or $M+N<K$ or things like that, we have to prove it for any $M,N$.

Yet partial sum is finite and $|s|^{m+n}=|s|^n|s|^m$ so we can separte summations.

$\displaystyle S(N,M)=\sum\limits_{n=0}^{N}|s|^n\sum\limits_{m=0}^{M}|s|^{m}=(\frac{1-|s|^{N+1}}{1-|s|})(\frac{1-|s|^{M+1}}{1-|s|})\to\frac{1}{(1-|s|)^2}$ for $|s|<1$.

The limit is independant of the way $N,M$ are going to infinity so the double serie is convergent.