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In the identity $$ \cos \left( \sum_i \theta_i \right) = \sum_{\text{even }n\ge0} (-1)^{n/2} \sum_{|I|=n} \prod_{i\in I} \sin\theta_i \prod_{i\not\in I}\cos\theta_i $$ one can prove the case of finitely many values of $i$ by induction on the number of such values, and questions of convergence are easy to treat when there are infinitely many (and similarly with sines).

In the identity $$ \tan \left( \sum_i \theta_i \right) = \frac{e_1-e_3+e_5-\cdots}{e_0 - e_2 + e_4 -\cdots} $$ where $e_k$ is the $k$th-degree elementary symmetric polynomial in the variables $\tan\theta_i$, the finite case is similarly routine.

What is known about convergence in the infinite case?

LATER EDIT:

I derived this odd identity that I have not seen elsewhere (so attribute it to me if you mention it in a publication, unless you find it in something earlier):

$$ \csc\left( \sum_{i=1}^n \theta_i \right) = \frac{(-1)^{\lfloor(n-1)/2\rfloor}(\csc\theta_1\cdots\cdots\csc\theta_n)}{f_{(n\operatorname{mod} 2)} - f_{(n\operatorname{mod} 2)+2} + f_{(n\operatorname{mod} 2)+4} - \cdots\cdots} $$ where

  • $f_k$ is the $k$th-degree elemenary symmetric polynomial in the variables $\cot\theta_i$
  • $\lfloor a\rfloor$ is the greatest integer $\le a$
  • $(n\operatorname{mod} 2)$ is the remainder on division of $n$ by $2$

so that the $\pm$ in the numerator is $$ \begin{cases} + & \text{if $n=1$ or $2$} \\ - & \text{if $n=3$ or $4$} \\ + & \text{if $n=5$ or $6$} \\ - & \text{if $n=7$ or $8$} \\ & \text{etc.} \end{cases} $$ A funny thing about this is that to get the case $n-1$ from the case $n$, you would presumably just set $\theta_n=0$, but then the cosecant and the cotangent both blow up. So you apply L'Hopital's rule, and fully half of the terms in the denominator vanish, if viewed as terms within $f_k$.

Can anything sensible be said about $\csc\left(\sum_{i=1}^\infty \theta_i \right)$?

And (also my own) $$ \cot\left(\sum_{i=1}^n \theta_i\right) = (-1)^{n+1} \left( \frac{f_1-f_3+f_5-\cdots}{f_0-f_2+f_4-\cdots} \right)^{(-1)^{n+1}}. $$ so we have $\text{even}\leftrightarrow\text{odd}$ alternation between the numerator and the denominator every time $n$ is incremented by $1$. Similar remarks about L'Hopital apply, and the same question about infinite sums can be asked.

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1 Answer 1

Alright I've figured out the answer to the first question I asked above.

Suppose $\theta_1+\theta_2+\theta_3+\cdots$ is an absolutely convergent infinite series.

The function $e_k$ is the $k$th-degree elementary symmetric polynomial in $\tan\theta_1, \tan\theta_2, \tan\theta_3, \ldots$,

Above I stated an identity one side of which is the cosecant of a sum. But there is another, far more well behaved expression that we can put on the other side of that identity, and we'll need it here: $$ \csc\left(\sum_{i=1}^\infty \theta_i\right) = \frac{\displaystyle\prod_{i=1}^\infty \sec\theta_i}{\displaystyle e_1-e_3+e_5-e_7+\cdots}. $$ To prove this, first do the case where only finitely many $\theta$ are non-zero, by induction on the number of such $\theta$, then it's pretty easy to think about convergence. In the same way, we get $$ \sec\left(\sum_{i=1}^\infty \theta_i\right) = \frac{\displaystyle\prod_{i=1}^\infty \sec\theta_i}{\displaystyle e_0-e_2+e_4-e_6+\cdots}. $$ (Just even indices rather than odd.)

So $$ e_1-e_3+e_5-e_7+\cdots = \frac{\displaystyle\prod_{i=1}^\infty \sec\theta_i}{\displaystyle\csc\left(\sum_{i=1}^\infty \theta_i\right)} $$ So the expression on the left actually converges if the one on the right does. What this means is that we need to think about $$ \lim_{n\to\infty} \frac{\displaystyle\prod_{i=1}^n \sec\theta_i}{\displaystyle\csc\left(\sum_{i=1}^n \theta_i\right)}. $$ The hardest part of this is the numerator and that's not so hard. We have $$ 1\le\sec\theta \le 1+\theta^2\text{ for $\theta$ small enough}. $$ We can use that to show the product in the numerator converges.

In going from the case of $n$ non-zero $\theta$ to $n+1$ of them, we don't just increment the number of $e_k$ that appear in the sum, but also every $e_k$ acquires more terms. But that is not hard to deal with.

So $e_1-e_3+e_5-e_7+\cdots$ converges!

And similarly for $e_0-e_2+e_4-e_6+\cdots$.

And from there it's not hard to show that the identity for the tangent of a sum still holds if the sum is an absolutely convergent series.

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