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For specific functions and $a_n >0$, we can say that $\sum a_n$ converges $\iff$ $\sum f(a_n)$ converges. I want to show this specifically for the case that $f(x)=sin(x)$. This is what I have thus far (which I admit is essentially nothing). Any direction would be welcome.

Proof:

$(\rightarrow)$ $\sum a_n$ converges. We know that:

(i) the partial sums of $a_n$ are bounded.

$(\leftarrow)$ $\sum f(a_n)$ converges. We know that:

...

This is not a duplicate of a previous question that I asked. Before, I asked for a proof of this concept (if it could be proved). This, I am asking for help with a direct application of the concept.

2 Answers2

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This is not true. For example, let $a_n=\pi$.

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I will prove this in one direction in the case $f(x)=\sin(x)$ and $a_n\in\mathbb R$. Also, I will assume absolute convergence for the $a_n$ series.

Assume that $$\sum_{n=1}^\infty |a_n|<\infty.$$ We know that $|a_n|\geq|\sin(a_n)|$ for all $n$, so $$\sum_{n=1}^\infty|\sin(a_n)|\leq\sum_{n=1}^\infty|a_n|<\infty.$$

In general, $\sum |a_n|<\infty\Rightarrow\sum |f(a_n)|<\infty$ as long as $f$ is lipschitz and $f(0)=0$.

The converse is true if $f$ is Lipschitz and if $0$ is the only root of $f$.

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